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Man is the only animal that blushes. Or need to. - Mark Twain, Pudd’nhead Wilson’s New Calender

Don’t go around saying the world owes you a living. The world owes you nothing, it was here first. – Mark Twain

Slipping between thousands, millions, billions and trillions without stopping to gasp in proper reverence for the sheer enormity of these relative sizes is like witnessing someone from the average family listen, simply nodding, as the president discusses a 7 trillion dollar federal debt and that average family’s income is less than $ 50,000 per year. And then it is too late. In fact, most people do just this, they slip seamlessly between huge numbers without the sobering knowledge of the relative sizes they are discussing. Over 500 billion cigarettes are smoked yearly in this country. A ¼ of a million people die each day on this planet. While you can see over 2 thousand stars on a moonless night, there are some 100 billion stars in our galaxy alone, the nearest of which is about 26.5 trillion miles away from our sun, the brightest of which is about 1,000 trillion times brighter than our sun. The 45,000 American soldiers that died during the Vietnam War is approximately equal to the number of fatalities from motor vehicle accidents in this country each year. There are nearly 295 million United States citizens; the over 200 million privately owned firearms owned in the United States comes to nearly 2 guns for every 3 U.S. citizens; the nearly one-third of a trillion dollars of the Annual Defense Department budget comes to nearly 1 thousand dollars for every U.S. citizen. Over 50 trillion pounds of TNT equivalence of the world’s nuclear weapons co-exist on this planet with more than 6.3 billion people, over 8,000 pounds for every man, woman and child. There are roughly 2 million US millionaires, while 35.8 million US citizens live

1 below the poverty level. World-wide, there are 7.3 million millionaires with a net worth of 27.2 trillion dollars, while over 1 billion people earn less than 1 dollar per day.

Let’s take any one of these statistics and mathematically think it through to its logical conclusion. Bear in mind, what we call mathematical, you may call intuition. It’s a technique you and I use every day. With a little mathematical approximation, we will see if it makes sense to accept the validity of this statistic that sadly 45,000 people die in traffics accidents each year. For starters, to make the estimation quick, let us begin by saying there are 50 states, thus 45000/50 gives us roughly 900 traffic fatalities per state yearly. Certainly, states with larger populations like California are prone to more accidents than the less populated states, like Wyoming. For the sake of expediency and ease of mental estimations, we can assume 900 traffic fatalities per year for each state. There are 365 days in a year, and if we average the 900 fatalities per year over the number of days in a year, we have 900/365, which is approximately 2 ½ . What does 2 ½ represent? For the average state, between two to three people are killed due to traffic accidents each day. Does this seem reasonable? Again, on any given day, for the larger states there may be more accidents and for the smaller states less. During holiday season or when there are severe accidents, more. The sad fact is, 45,000 traffic fatalities per year, unfortunately does seem accurate.

All of us live in a society where numbers are meaningful, yet the masses are not educated to understand the true meaning of numbers. How apropos if everyone was simply numerate, literate, fluent in understanding the numbers that abound us. Empowered to critique, understand and take charge of our own conclusions with respect to the numbers we are confronted with everyday. We live in a numerate world, we need to be numerate citizens. The sole goal of this book is for you to develop the mathematical tools to be mathematically literate in our numerate society.

To be utterly honest, this is not a problem unique to mathematics. The vastness of time is to the geologist as the enormity of space is to the astronomer. For most of us, comprehending the sheer size of space or vastness of time seems mind boggling, and at first glance it may appear to be something beyond our comprehension. But let us take a moment to examine this wonderfully seemingly limitless world around us.

Geology: Vastness of time Five thousand years of recorded human history seems like an awfully long time, but recall, humans first appeared nearly two million years ago. When we are talking about geologic processes, this amount of time is a mere blip on the geologic time scale. The earth itself is thought to be 3000 times older than when man’s first footsteps plodded forth.

Do you recognize the relative sizes of numbers like 5 billion, 700 million, 200 million, 2000, and 400? Sure, you can order them, but how much bigger is 5 billion than 700 million, or 700 million to 2000? In other words, do you have an ‘intuitive feel’ for the relative sizes of these numbers. Geologists have been battling this very crusade of trying to educate the public into realizing the comparative magnitude of geologic time. The

2 most popular tool they employ is the comparison technique. They begin with comparing the earth’s 5 billion year age to one week’s worth of time.

Let’s compare. Let’s assume one week is the age of the earth. Starting with Monday at 12:00:01 am, the other celestial bodies are singing “Happy Birthday” to the earth. So, on this time line, present day (right now) is Sunday night, midnight. Well, from Monday through Saturday, very little of anything noteworthy occurred. To be honest, very little of the earth’s formational processes prior to evidence of life routinely makes the evening news or the front page, but a mere 700 million years ago, the first fossil animals appeared. Let’s ask ourselves, when did this “700 million years ago” occur in the course of our week? Fractions (which students and most members of society try to avoid) now become our ally. We are going to look at the ratio of what 700 million years is to 5 billion years and then we will know what proportion of time is equivalent to our one week time period. That period of time we will remove from the end of our week.

So, if 1 week is to 5 billion years, then how much of that week is to 700 million years? We will estimate 700 million out of 5 billion. Working with the ratio 700,000,000/5,000,000,000 is preposterous to deal with. We will reduce this ratio to 700/5000 or 7/50. We can use 7/49 for a mental estimation, or 1/7 of a week, which translates to one day of the week. Early, very early Sunday morning, the last day of the week, the first fossil evidence of life appears on earth. See the power of estimation! Let’s continue.

Some 220 million years ago, the first dinosaurs appeared. Using our estimation technique again with respect to the age of the earth, we construct a ratio 220/5000. We can approximate this as 200/5000 = 2/50=1/25. OK, now what? We need to know how much time corresponds to 1/25 of a week and remove it from the end of the week. Well, 7 days divided into 25 parts is about 1 day for every 3 ½ parts or 24 hours for every 3 ½ parts or about 1 part per 7 hours. Thus, 7 hours before midnight, or around 5 pm Sunday evening, the dinosaurs first appeared.

Next, let’s tackle the number 2,000 from our introductory paragraph. The Birth of Christ was roughly 2,000 years ago. The question before us, in the development of our timeline, if the earth’s age compressed to a week, where is the Birth of Christ located on the timeline? We continue by reiterating the question, what is 2000/5,000,000,000 of a week? Or more specifically, what proportion of time is the ratio of 2000 years to 5 billion years with respect to one week? We will simplify the ratio by dividing 1,000 into the numerator and denominator, and re-ask the question, what amount of time is two-five millionths (2/5,000,000) of a week?

This is as an unfriendly a ratio as you can imagine, so we will enhance our comparison method by using a method chemists employ. Chemists use the ‘mole method’ or ‘stoichiometry’, mathematicians use a similar method to tackle unit conversions.

First, let’s firm up our language, specifically, what amount of time is 2/5,000,000 of a week ago? Including the units with the ratios, we have:

3 2 7 days weeks x  0.0000028 days ago. 5,000,000 1week Notice, the ‘weeks’ unit cancelled. In the process of providing a more familiar context for this small amount of time, we continue with the unit conversion. Now we have: 2 7 days 24 hours 60 min utes 60 seconds weeks x x x x  0.241 seconds or 5,000,000 1 week 1 day 1 hour 1 min ute about ¼ of a second ago. What does this ¼ of a second represent? If the earth’s age was compressed to a week, the Birth of Christ would have occurred within the last ¼ of a second. This means, the birth of Christ and any latter events, such as the wisdoms of Budha, works of Shakespeare, Beethoven or Picasso, arrival of the Pilgrims to Plymouth Rock, Revolutionary war, Babe Ruth’s 60th homer or when ex-vice president Al Gore invented the Internet, all would have occurred within time it took you to read this word, under a ¼ of a second ago.

All right, so with the use of setting ratios equal to each other, which in mathematics we recognize as proportions, we can get a feel for the relative magnitude of any numbers we encounter. If the numbers represent size, distance or some magnitude, the comparison and estimation method provides a practical tool that we can easily apply. Abstractly, we have just used the comparison method on a timeline. But, this comparison method is no different that placing the following visual in your mind: Picture the earth, a rock too big too comprehend. Imagine it in all it’s glory, picturesque, scenic, it’s soaring mountain ranges, it’s cascading valleys and vast oceans. Now, imagine shrinking it down to the size of a cue ball. This cue ball would be smoother than any cue ball you could ever construct. That is how large the earth is compared to the mountains and valleys you see.

Exercise Set:

For problems 1 – 4,use the same breaks 2. Compare the earth’s age to 1 year in time we used above and recall, the earth is 5 billion years old. 3. Compare the earth’s age to 18 years A. 700 million years ago, the first fossil (the age of our average freshman) animals appeared. B. 220 million years ago, the first 4. Compare the earth’s age to a century dinosaurs appeared. C. 2000 years ago, The Birth of Christ For problems 5 to 9, imagine the D. roughly 400 years ago, the writings celestial body below to be roughly the of Shakespeare shape of a sphere, who’s circumference Construct a timeline locating each item is given by the formula C  2r . If a above on the time frame below, as we ribbon was tied around the sphere in its did in this text, for each of the following: middle, how long would the ribbon have to be? If a second ribbon was extended 1. Compare the earth’s age to 1 month six feet from the object (sphere), how (assume a 30 day month)

4 much longer would the new ribbon have 7. Jupiter, whose diameter is 88,700 to be? miles 5. baseball, whose diameter is 2 ½ 8. the sun, whose diameter is 864,950 inches miles

6. earth, whose diameter is 7,927 miles 9. discuss the similarities of the answers above. What have you learned?

Astronomy: Enormity of space

Let’s turn our attention to the enormity of numbers from the size of the earth and it’s fellow celestial bodies in contemplating the relative size of some astronomical record holders.

 The Sun’s diameter is 864,950 miles  The Largest Planet is Jupiter, whose volume is over a thousand times that of the earth.  The Smallest Planet is Pluto, which is 0.0017 as big as the Earth  The Largest Star known is R136A in Large Magellanic Cloud, which is 400 to 1000 times the Sun’s mass  The Most distant object is Quasar Q000-26, which is 18,000,000,000 (or 18 billion) light years from the earth.

In reality, these numbers are relatively pleasant magnitudes. Quick, try this. Which is smaller? 0.00000000007 or 0.00000000017? Quick, which is bigger? 1800000000000000000000000000 or 1080000000000000000000000000. Think these numbers are contrived? Are you thinking we might as well as asked how many angels could fit on the tip of a pin? The smallest organism of any kind is a viroid, a virus like plant pathogen. These little creatures are 0.00000000007 inches in diameter. The sun’s weight is 1,800,000,000,000,000,000,000,000,000 tons. The number of angels that could fit on the tip of a pin is answerable if you know the details as to the size of the angel and the diameter of the pin’s tip.

For most of us, when we first see numbers such as those representing the diameter of a viroid and the weight of the sun, we become mesmerized, paralyzed, and can not think. It is like being confronted by the weaving head of a cobra. Even mathematicians balk at using numbers with so many digits. Writing numbers with 28 decimal places becomes both tedious (too many zeroes) and impractical because we can not immediately tell the relative size of the numbers. And this is where scientific notation comes in. In mathematics and the related sciences, scientific notation is used to represent really big or really small numbers. The power of scientific notation is the immediate clarity of the size of the number, written succinctly.

5 Scientific Notation A positive number is written as the product of a number greater than or equal to one but less than ten multiplied to a integer power of ten.

Thus, to represent the sun’s weight of 1,800,000,000,000,000,000,000,000,000 tons, we will rewrite this number with just one nonzero digit to the left of the decimal point multiplied by an integer Using scientific notation this unmanageable 27 digit long number is succinctly rewritten as 1.8 x 1027.

Similarly, negative exponents on the base of ten are used to indicate small numbers. That is, 0.00000000007 inches is rewritten in scientific notation is 7.0 x 10 - 11.

Why do we need such representations for large and small numbers? Answering these questions is how basic numeracy help us to understand the world around us. Why does the typical college graduate need to know large and small numbers? Where are we confronted by this in every day life? Ask yourself how much money per person the 7 trillion dollar federal debt comes to per person if there are about 290 million United States citizens. Ask yourself, if some one knew your user name for your hot mail account, how many passwords would the person need to go through if they were arbitrarily trying to enter your site before they were guaranteed access? We will actually see this huge number later, and we promise you, it is larger than you think.

Let’s try an exercise and restrict ourselves to our solar system.

Table 1. Our solar system, excluding the moons, satellites and other members. Table 1 Body Body diameter in miles Orbit diameter in miles Sun 864,950 Mercury 3,000 72,000,000 Venus 7,700 134,400,000 Earth 7,927 186,000,000 Mars 4,219 282,200,000 Jupiter 88,700 986,600,000 Saturn 75,100 1,772,000,000 Uranus 29,300 3,560,000,000 Neptune 31,200 5,580,000,000 Pluto 3,700 7,340,000,000

Before we proceed with this exercise, let’s just take a moment to glance at the numbers in the table above and think about them. We’re not so busy that we cannot pause for a little reflection. Scan the above table, and think about the relative size of the planets and their orbits. Now, close your eyes and recall the last time you were in a planetarium. Does something strike you as odd? If not, numeracy has not yet set in. There is a great line in Rocky III, where Apollo Creed turns to Rocky at the end of the movie and says, “you fight great, but I am a great fighter.” Well, if what we have said up until now, has made perfect sense to you, but if the planetarium you envisioned doesn’t seem odd, then you think great, but let’s make you a great thinker.

6 We have a lofty goal, to read Table 1 and from it, put in context the precise size of the solar system. Now, what does it mean to contextualize the precise size of the solar system? Two visualizations must float through our minds. First, How large are the planets relative to each other? And second, how far apart are the planets from one another? For example, when we read Table 1, if we close our eyes, can we see a sphere whose diameter is 865,000 miles? If we could comprehend the enormity that is the size of the solar system, we would not need to use the comparison method to contextualize the solar system. The comparison method gives us the power to see numbers in the million and billions in place them in terms we can visualize. We can’t imagine the sun’s true size, however, we can imagine a scaled down version of the sun.

Guy Ottewell, in his THE THOUSAND-YARD MODEL, OR THE EARTH AS A PEPPERCORN illustrated this very notion. Succinctly and clearly, he provides us a visualization as clear as a icy mountain lake in the form of a wonderful children’s activity that went something like this. First, let’s look at the table displaying the planets’ diameters and distances traveled in their orbits. Now we turn to the Comparison Method. Let’s begin with the sun whose diameter is approximately 865,000 miles. Can you visualize 864,950 miles? Neither can we. So, let’s reduce it to 8 ½ inches, scaling down our solar system so the scale is 100,000 miles to an inch. This is a good scale to start with, because although we can not visualize 864,950 miles, we can visualize a soccer ball that has a diameter of about 8 or 9 inches.

Table 2. Clearly, if our scale was 100,000 miles for every inch, Mercury’s body diameter would be 3000/100000 or 0.03inches. Below is a new table with the old and new measurements side by side for comparison purposes. Body Body diameter before and after the scale Miles  Inches Sun 864,950  8.50 Mercury 3,000  0.03 Venus 7,700  0.08 Earth 7,927  0.08 Mars 4,219  0.04 Jupiter 88,700  0.89 Saturn 75,100  0.75 Uranus 29,300  0.29 Neptune 31,200  0.30 Pluto 3,700  0.03

Now, in relative terms, if the sun was a soccer bowl, the other planets’ sizes would be:  Mercury: a grain of sand  Saturn: a medium gum ball  Venus: a Bee-Bee  Uranus: a tiny marble  Earth: a Bee-Bee  Neptune: a tiny peanut  Mars: a wild flower seed  Pluto: a grain of sand  Jupiter: a large gum ball

7 All of these items you can give to young children to place in the specific location around the soccer ball to represent the solar system. This was the easy part. Here is where the notion of an indoor planetarium falls apart.

Because the next question is “how much space is needed between the planets?’’ If 36 inches (one yard or one pace walking) represents 3,600,000 miles, the distance from the sun to the first planet, mercury, would be approximated by considering half of the diameter of the orbit. The diameter of mercury’s orbit is 72,000,000 miles or on our scale, 20 yards. Why? (yes, it’s rhetorical.) This means the distance from the sun to mercury is approximately 10 yards on our scale. Okay, we go outside.

Table 3. Using this same scale of 100,000 miles for every inch, our solar system excluding the moons, satellites and other members, would have the following scaled back dimensions: Scaled down Scaled down Body body diameter distance from the sun in inches in paces Sun 8 ½ Mercury 0.03 10 Venus 0.08 19 Earth 0.08 26 Mars 0.04 39 Jupiter 0.89 137 Saturn 0.75 246 Uranus 0.29 494 Neptune 0.30 775 Pluto 0.03 1019

Picture a child reconstructing a model of the solar system. Setting a soccer ball down in the center of a large field, she will step these 10 paces and mark Mercury with a grain of sand. She steps another 9 paces, and she places the Bee-Bee. Steps another 7 paces, another Bee-Bee, aka the earth; another 13 paces, a wild flower seed; another 98 paces, the large gum ball; another 109 paces (or more than a football field), the medium gum ball; another 248 paces or two and a half football fields, a tiny marble; another 281 paces or almost three football fields, another tiny marble; and finally another 244 paces or two and a half football fields, a grain of sand representing is placed more than a half a mile from the soccer ball. Once the grain of sand drops from lil’ Nina’s hand, she would naturally turn to see the original soccer ball. Good luck lil’ Nina.

To get straight to the point, any solar system model that you have gazed at on the ceiling of a planetarium was grossly undersized. The notion any indoor planetarium can adequately represent our solar system on any reasonable scale is preposterous. Spatially, the planets are far too small compared to distances, which are far too vast.

Now, consider that the orbits do not stay in a straight line, rather they move in relatively elliptical paths at differing speeds, and this means usually, the planets are even further

8 apart than this what we just did when we lined up the planets along a straight line. In short, the relative distances on the above table need to be appreciated in a manner other than the assumption that the planets are always co-linear. Rather, the distances are always changing and this needs to be recalled as we observe the unusualness of such results as the that the earth is only 1/40th of the way to Pluto from the Sun or that the planet with an elliptical path closest to the halfway point from the Sun to Pluto is not one of the planets in the middle, but rather Uranus.

Now that we have a little better understanding of the vast distances in our solar system, let’s investigate the plausibility of alien UFO’s or visitors from other solar systems. The model previously discussed is just our solar system, yet there are 1.0 x 1011, that is 100,000,000,000 or 100 billion stars, in our galaxy alone. And how many galaxies are there?

When contemplating the notion of the existence of visitors from outer space, we need to factor in the vastness of space and the enormity of time. We are truly a mere spec of dust floating in the vastness that is space during an instant in the enormity that is time. Life happens to exits now on this planet. For aliens to visit us, several things must occur. First, they must be alive during the same interval of time on the grand time scale that humans are alive and well on this planet. Second, they must be able to traverse vast distances to find our singular spec of dust. Third, what would be their motivation to arrive at our spec of dust? What makes us so interesting? It is an assumption to believe aliens have the same desires to meet or find new life on a remote planet other than their own.

The closest star to our Sun is Alpha Centauri, which is about 4.5 light years away. And since a light year is approximately 6 trillion miles, this nearest star is about 26.5 trillion miles away from our sun, much less our earth, What is the chance that aliens would visit us from this star’s system, or even find us at all? The sun, after all, is only one of hundreds of billion of stars in one little ol’ galaxy, the Milky Way galaxy. And the Milky Way galaxy is one of hundreds of billions of galaxies in the universe. What would motivate them to choose our galaxy? Or sun? Our planet? Even if they did find us, what is the chance they would find us now? Fat chance. The precise odds, you ask? I am sorry, that is in Chapter Two. But we guarantee you, lots of coincidences would have to come together for our third rock from the sun to be visited by beings from outer space.

We can gain an even better understanding as to the precise size of our solar system if we refine our comparison method to talk in terms of the sizes of the Sun and planets relative to the Earth’s size or the planets distance from the Sun relative to the Earth’s distance from the Sun. Think about Table 1, which has the exact measurements, numbers that are generally regarded too large to comprehend in visual manner. Now, think about Table3, the table where the scaled down version of measurements appear, numbers that are easier to visualize, but they keep the same proportion for the measurement they represent. If we asked you how much more space does Jupiter occupy relative to the Earth or perhaps how smaller is Pluto than the Earth or even how many times farther is Pluto from the Sun than the Earth, which table would you use? Table, 1 or Table 3?

9 So, let’s do just this, further refine our understanding of the size of our universe and discuss the size of the Sun and the planets relative to the size of the Earth. For example, ‘how much larger is the Sun than the Earth?’ Would you believe the Sun is over a million times larger than the Earth. To show this, we will use Table 3 because this scaled down version has numbers we can easily both comprehend and work with. This Table is in proportion to the true celestial measurements, with the proportions well preserved. We will assume each of our celestial body is a sphere (remember the earlier cue ball 4 analogy), whose volume could thus be calculated by V= p r3 , where r is the radius of 3 the planet. Look at Table 4 below. This table compares the celestial body’s diameter and volume to the earth’s diameter and volume.

Since we are comparing the diameter of to Sun to the diameter of the Earth, we form a ratio, and place the Earth’s diameter in the denominator. The Sun’s diameter is over two hundred times larger than the Earth’s diameter as seen in ratio of the two diameters: Diameter of the Sun 8.5   212.5 Diameter of the Earth 0.04

And, the Sun occupies over a million times the space the Earth occupies, as seen in the 4  4.253 3 Volume of the Sun 3 4.25 ratio of the two volumes:   1,194,463 Volume of the Earth 4  0.043 0.043 3

Before we proceed to complete these columns of our Table 4 below, let’s take a moment to check the status of your numeracy. Please provide estimations for what you believe are the logical relative sizes of each body when compared to our Earth. Use the table of your choice, either Table 1 or Table 3 to guide you. Once finished with the estimation, complete the Table 4 below.

Table 4. Body Diameter of Body Volume of Body relative to the Earth relative to the Earth Sun 212.5 1,194,463 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto To see the complete table, work Problem 4 in the Exercise Set for this section.

10 Problem One - Scientific Notation

A 60 year old woman has lived over 1892000000 seconds. What is her age in seconds expressed in scientific notation?

Solution

First, rewrite 1892000000 with the appropriate commas, 1,892,000,000. It is convention that we rewrite this number with just one nonzero digit tot he left of the decimal point multiplied to an integral power of teen. So, rewrite 1,892,000,000 as the product of a number greater than or equal to one but less than ten, 1.892 multiplied to a integer power of ten, 109, because we are essentially counting the number of place values right of the 1. Thus the answer is 1.892 x 109 seconds.

Problem Two - Scientific Notation

A serving of diet ice tea has 1.2 mg of sodium, which is 1.2 x 10-3 grams of sodium. What is another representation of this measure? a) .012 g b) .0012 g c) .00012 g d) .000012 g

Solution

In scientific notation, negative exponents on the base of ten are used to indicate small numbers. So, 1.2 x 10-3 can be thought of sliding the decimal point to the left three places. The answer is .0012 g; choice b).

Time and Again, Number Sense

To help give us a better sense for the relative size of numbers, we will visualize the amount of time a stated number of seconds truly is:

 Ten (10) seconds is the time takes to dial a long distance number or do 9 push- ups. 2  One hundred (100) seconds is 1 minutes or a little less than 2 minutes, which is 3 the time it takes to listen to little less than half of your favorite song. 2  One thousand (1,000) seconds is 16 minutes or the time it takes to drive 16 3 miles on a highway. 2  Ten thousand (10,000) seconds is a little more than 2 hours or a little longer 3 than 2 hours and 40 minutes, which is the time it takes to watch the average baseball game.

11  Hundred thousand (100,000) seconds is 1 day, 3 hours, 46 minutes and 5 seconds, which is a little longer than your one day, like your own birthday or New Year’s Eve day, the one day where we can be awake for 24 hours (or 27 actually).  One million (1,000,000) seconds is 11.5 days or a little less than the 12 days of Christmas.  One billion (1,000,000,000) seconds is 32 years. 32 years is the amount of time that has passed since Richard Nixon visited China, The movie The Godfather was released, The Allman Brothers put out Each A Peach, The Munich games saw Isreali athletes massacred, Idi Amin forced out all Asians from Uganda, George Wallace was shot and Nike put out its first running shoes.  One trillion (1,000,000,000,000) seconds ago is when the prominent brow, jutting jaw, large front tooth Neanderthal man died out.

One last qualifier so we may avoid any unnecessary confusion that actually does exist on opposite ends of the pond. In the United States of America, we would say the population of the world growing at a tremendous rate, it is nearly 6.4 billion, while in England we would have to amend our voice to say the population of the world is nearly 6.4 thousand million. This is because every group of three zeros has a name in the United States, but in England, a unique name is given to every group of six zeros. Both countries are in numerical alignment until we reach the billions place. One billion is written as 1,000,000,000 in the USA, but is written 1,000,000,000,000 across the pond. See the table below for clarification.

Name United States of America England Billion 1,000,000,000 1,000,000,000,000 Trillion 1,000,000,000,000 1,000,000,000,000,000,000 Quadrillio 1,000,000,000,000,000 1,000,000,000,000,000,000,000,00 n 0 Quintillion 1,000,000,000,000,000,000 1x1030 Sextillion 1,000,000,000,000,000,000,000 1x1036 Septillion 1,000,000,000,000,000,000,000,000 1x1042 Octillion 1,000,000,000,000,000,000,000,000,0 1x1048 00

12 Exercise Set

1. Our solar system, excluding the moons, satellites and other members. 3. Fill in the table below. Fill in the table. Body Diameter of Volume of Body Relative Relative Body Body distance from distance from relative to the Relative to the the Sun when the Sun when smallest planet smallest compared to compared to planet the Earth’s Mercury’s Sun distance from distance from Mercury the Sun the Sun Venus Sun Earth Mercury Mars Venus Jupiter Earth Saturn Mars Uranus Jupiter Neptune Saturn Pluto Uranus Neptune Pluto 4. Complete the table below: 2. Let’s assume that even though each Body Diameter of Volume of Body Body orbit is an ellipse, we will reduce the relative to the Relative to the orbit to roughly a circle, so that the Earth Earth distance traveled by the planet can be Sun more quickly approximated by taking Mercury Venus the circumference of the circle, C= 2p r , Earth where r is the radius of the orbit. Mars Our solar system, excluding the moons, Jupiter Saturn satellites and other members Uranus Body Relative Relative Neptune distance of the distance of the Pluto planet’s orbit planet’s orbit when when compared to compared to 5. PROJECT: Access through the Internet the distance of Mercury’s three maps, each with their own scale: a map the Earth’s orbit of the United States, a map of the world, and orbit Sun a map of our solar system. Then, using your Mercury school at the starting point, determine what is Venus located ten, hundred, thousand, ten thousand, Earth hundred thousand , million, billion and a Mars Jupiter trillion inches away? Saturn Uranus Neptune Pluto

13 7. Understanding reasonableness of orders of 6. Understanding orders of magnitude. magnitude from the world of science. Most of us do not have these trivial facts Most of us do not have these trivial facts below memorized, but we should be able below from the world of science memorized, to reasonably estimate their order of but we should be able to reasonably estimate magnitude. In a comparative sense, their order of magnitude. Match the number these orders of magnitude each have below with the most logical measurement. differentiating aspects to them. Match Each number represents a different order of the number below with the most logical magnitude, a different power of ten. measurement. Each number represents a different order of magnitude, a different a. 100,000 1. Number of power of ten. permanent teeth in an a. 294,000,000 1. Population of adult that are used to the US, 2004 chop food into bits b. 1.3 x 107 2. Population of b. 68 2. Number of years the earth, 2004 ago that life developed c. 41,994 3. Land area of c. 36,640 3. Number of years US in square ago that modern miles humans first existed d. 6,300,000,000 4. Earth’s width, d. 32 4. Oldest recorded in miles age of an animal (a e. 2,000 5. Distance from clam) in years the sun to earth, e. 22,834 5. In mph, the speed in miles of the fastest swimmer f. 92,955,820 6. Earth’s width, (cosmopolitan in yards sailfish) g. 1.1 x 1021 7. Earth’s surface f. 6. In feet, the height area, in m2 225,000,000 of the average h. 3,537,438 8. Earth’s orangutan volume, in m3 g. 7. Deepest part of the i. 0.5 x 1015 9. Average. 600,000,000 ocean, the Marianas household Trench, in feet income, $, in US h. 220 8. Height of the 1999 highest volcano, j. 107,501 10. Number of located in the stars you can see Argentine Andes, in on a moonless feet night i. 4 ½ 9. Depth, in feet, of k. 8,000 11. Seating the largest land capacity of canyon, the Grand University of Canyon Michigan’s j. 5,300 10. Years ago the football stadium oldest dinosaur roamed the earth

14 8. Understanding reasonableness of 9. Exploring order of magnitude: orders of magnitude from the world of Discovering the difference between geography. Most of us do not have these arithmetic versus exponential growth. trivial facts below from the world of Something to consider: Most biologists agree geography memorized, but we should be that while our food supply grows able to reasonably estimate their order of arithmetically, our population grows magnitude. Match the number below exponentially. The rapid increase in our with the most logical measurement. population’s exponential growth also in turn Each number represents a different order explains why biologists also inform us that of magnitude, a different power of ten. right now 10 to 20 percent of all the people who ever lived are alive right now. You may a. 105 1. In feet, elevation of be asking yourself, is there a question in all Mt. McKinley in of this? Yes. Do you really see how sharp Alaska the difference is between the two types of b. 70 2. In nautical miles, growth? This example will help us see the the length of the difference in growth rate clear as day. Use Mississippi River from your calculator any way you would like to Minnesota to the Gulf answer the below two questions. of Mexico a) Arithmetic Growth: How many times do c. 707,400 3. In degrees, average you need to add 2 to itself to equal or exceed temperature in one hundred, one thousand, one million, one Phoenix, Arizona in billion and one trillion? July b) Exponential Growth: How many times do d. 8 4. In square miles, the you need to multiply 2 to itself to equal or area of the continent exceed one hundred, one thousand, one of Asia million, one billion and one trillion? e. 2,348 5. Percent of the Answer this question by constructing a table, world’s surface labeling the rows one hundred, one thousand, covered by oceans one million, one billion and one trillion and f. 6. Volume in cubic labeling the two columns arithmetic growth 1,000,000,00 yards of the world’s and exponential growth. Place the growth 0,000,000 largest dam, Syncrude factor (the answer to our question) on the Tailings, in Alberta table. Look at the results. And then become g. 0.0017 7. Percent of the worried about our food supply. world’s population for Australia 10. How many times larger is England’s h. 20,320 8. Percent of the version of a billion compared to the United world’s population for States version? South America i. 0.5 9. Number of times 11. How many times larger is England’s brighter the brightest version of a trillion compared to the United object ever recorded States version? (Quasar S5 0014 + 81) is than the sun 12. How many times larger is England’s j. 17,240,000 10. Size of Pluto is version of an octillion compared to the ______as big as Earth United States version?

15 Set Theory

16 A set is a collection of objects and the objects are called elements of the set. The government forms sets each time they sort data accumulated from the 1040 tax form we all fill out annually. There are many 1040 tax forms to choose from. Consider tax form 1040.com above. The form is an Income Tax Return for Recently Laid-Off Dotcom Employees With No Job Prospects in 2000. Within the bubble to the left of where you enter your social security number are a series of statements which help you determine if you are an element of this set. To make such a determination, you must indicate you belong you belong to one of the distinguishing characteristics of the set: are you in the set of all people who were recently asked to place the contents of your office in a card board box and surrender your ID card or are you in the set of those people who were born after Jan. 1, 1973 or in the set of those people who hold a liberal arts degree or in the set of those who ride a scooter to work or a member of the set of people whose title started with ‘vice president for … ‘ Yes, this form is a satire created and posted on the web by Tim Carvell and Adam Horowitz (http://www.1040 Form - eCompany Version.htm). But, the concept of creating different sets that are determined by the true 1040 forms is not satiric. Data on all tax paying citizens is collected each year and then separated into distinct and well defined sets.

When we say a set is well-defined this means there is a clear cut method to determine whether or not a particular element belongs in the set. For example, let’s call set A the set of counting numbers less than five. It is easy to decide that the number 6 is not in set A, while the number 3 is in the set. This is what we mean by a well defined set. What is a set that is not well defined? If we define set B as the set of small counting numbers, this set is not well defined because it is impossible to determine if a given counting number is small because the word ‘small’ is so subjective. Is the set of vowels from the English alphabet a well-defined set?

There are many ways to notate a set. If the number of objects is palattable, and by pallatable I mean the set contains a small enough number of objects where viewing each object is not too cumbersome of a task, then we could list the objects separately, using commas to separate each object belonging to the set. A complete or implied listing of all elements in the set is known as roster notation. The sets A = {0, 1, 2, 3, 4} and B = {0, 1, 2, 3 … 400} are examples of roster notation used in defining sets with 5 and 401 elements respectively. The ellipsis, “ … “ is used to mean you fill in the missing elements in the obvious manner or pattern, as there are too many to actually list out on paper. We call the number of elements in a set the cardinal number of the set, we refer to it as the cardinality of the set and we denote this with the following notation, n(A). So, we have n(A) = 5 and n(B) = 401.

Set-builder notation is another way to define a set. In using set builder notation, each element of the set is identified in a well defined manner. We can describe set B above by saying all the counting numbers between 0 and 400. Using set builder notation, we write {x| x is a counting number between 0 and 400}. The vertical bar, “|”, is read as “such that” so this notation read aloud is `the set of all x such that x is a counting number between 0 and 400.’ To use set builder notation, there must be a discerninable pattern you can refer to when C ={p ,1 ,0.046, - 2} describing the elements in the set. For example, the set 5 is not a good set to describe with set builder notation. (Coming up with a pattern for describing the elements would be very difficult.) Set builder notation is often used when the roster method

17 is too cumbersome or even impossible because the set contains too many objects. Can you think of a set with too many objects? That is, to put it in the language of mathematics, we could say that if the set’s cardinality is too large, set builder notation is preferred given the description of the elements is short and sweet.

Now, let’s get away from boring old numbers and explore every day sets with every day people, like billionaires. Do the rich keep getting richer? In 2004, Forbes magazine published a list of the world’s billionaires. Though the extreme salaries of athletes and entertainers frequently grab the headlines, their multi-million dollar deals are dwarfed by the vast wealth of the wealthiest business men and women. In 2004, who were the richest of the rich, you ask? Heading the list, the top ten, were Microsoft's Bill Gates (USA) at $ 46.6 billion (up from 12.9 billion a mere ten years earlier), investor Warren Buffet (USA) at $ 42.9 billion (up from 10.7 billion ten years earlier), Karl Albrecht from Germany at $ 23 billion, Prince Alwaleed Bin Talal Alsau from Saudi Arabia at 21.5 billion, Microsoft’s Paul Allen (USA) at 21 billion, and then the Waltons (USA), Alice Helen, John and S Robson, at $ 20 billion a piece and the list goes on. You are probably wondering, “what’s the cardinality of the set containing all of the world’s billionaires in 2004.” We’ll get back to this in a moment.

In contrast to billionaires, according to the US Census Bureau, an average American family in 2003 had a median income worth $ 43,318. It would take a city of more than a million such families to equal the net worth of Bill Gates and amazingly, this figure is up from the quarter of a million families it would have needed to equal Bill Gates worth some ten years ago. Are the rich getting richer, we re-ask?

If we were to continue the list of billionaires by listing each billionaire separately, this would an example of roster notation. Writing the set out would be time consuming and monotonous. It would be easier and more elegant to use set builder notation because the cardinality for this set of billionaires in 2004 is 552. So, what does the notation to describe this set look like? A = {x|x is a billionaire in 2004} Is A well defined? You betcha! Either someone was a billionaire in 2004 or they were not. Were you in the set?

The notation x A indicates x belongs to the set A and is read “x is an element of set A.” If we need to state that an element does not belong to a particular set, the following notation, x B is read “x is not an element of set B.” Consider the following set given to us in roster notation, V = {a, e, i, o, u} and V can also be described using Set Builder notation, V = {x|x is a letter of the alphabet and always a vowel}. And y V . The subset relation A B states that every element of A is also an element of B. Furthermore, two sets A and B are equal if A B and B A. So, if B = {x| x is a millionaire}, then A B is correct because every billionaire is a millionaire (at least 1000 times over.) For another example, let D ={ Q | Q is a letter of the English alphabet} and V 虳 . What is n(D ) ?

The previous examples have all been mathematical conveniences to define sets, determine whether or not an element belongs to a set, or even if one set belongs to another. Now we will focus on combinging sets, but notice the word ‘combining’ is vaque. There are many

18 ways to combine sets. The union, denoted AU B , and intersection, denoted A∩ B operate to form new sets from previously defined sets by the following specific rules. The set AU B is defined to be the set of all elements that are either in A or in B or in both. The set A∩ B is the set of all elements that are in both sets A and B. In other words, AU B is the set that contains all of set A as well as all of the elements in set B. And A∩ B is the set containing only those elements in the overlap of both sets.

So, returning to set A = {x|x is a billionaire in 2004} and if set B = {x| x is a United States citizen in 2004}, then AU B is the set of all of the world’s billionaires as well as all of the citizens of the USA in 2004. We are in the set AU B and it does feel good to be thrown into a set along with the world’s billionaires. In fact, all of the 290,809,777 US citizens in Janueary of 2004 are in AU B as are all 552 world’s billionaires. Of course, there were 277 US billionaires, so there are 277 people who were counted twice. If we remove them, we have 290,809,777 + 275 non-US billionaires in the set AU B , giving us 290,810,052 elements in the set and n( AU B ) = 290,810,052. Good thing we are not using roster notation and instead we are using the more descriptive set builder notation. The set A∩ B is the set of American billionaires in 2004, of which there were 277 elements and probably most of us reading this are not one of them. And n( A∩ B ) = 277. Incidentally, neither was any athlete, despite the inflamatory headlines about how much money they earn. Now, if we are using sets to discuss people, let us specify which people we may draw from. In other words, we are not looking at billionaires that are deceased. We have implied but not said this outright. To say it outright, we would use the following language: Let our Universe be all living people in 2004. We would denote this set as U = {x|x is a person who is alive for all of 2004} and it is called the Universal set. Set A and set B are subsets of the Universal set U. Since most of us are not billionaires, we would belong to a set called the compliment of set A. The notation used to denote the compliment of set A is A and this set consists of all elements from the Universe which do not belong to set A. In using set B defined from above, we would describe B as the set containing all of the world’s population in 2004, except for the citizens of the United States. Let’s use the sets A and B to define this group of people, billionaires from 2004 who are not Americans or more succinctly, the foreign billionaires. In set notation, this is A minus B. This isolates all the billionaires who are not Americans. Many times in mathematics it is useful to form a set that would include all the elements from A that are not in B. The notation for this operation is A \ B, which is read A minus B or A not B. Let’s add two more sets to our discussion. Set C is defined as any person who was on welfare in 2004; C = {x|x is a person who was on welfare in 2004} and set D will be the set containing all people who were alteast once in their lives on welfare; D = {x|x is a person who once in their life was on welfare}. Notice, there is no overlap between the elements of sets A and C because no billionaire in 2004 was also on welfare that year. When no overlap exists between two sets, we use the word disjoint to describe these sets. Mathematically speaking, if A∩ C is empty, then the sets A and C are called disjoint and we would write A∩ C = . The symbol represents the empty set or null set. It is customary to use this

19 symbol to represent a set with no elements. Here is a fun question. Why are the sets A and D not disjoint? JK Rowling, the author of the Harry Potter books, was once on welfare. While on welfare, she wrote the first volume of the Harry Potter series. Today, she is a billionaire. Hence, the sets A and D are not disjoint, and moreover, n( A∩ D )= 1.

There are many types of real world sets some fascinating, some full of intrigue, some just plain boring. But, any collection of objects will suffice. From the world of music, musical set theory has many forms to it. The notion of defining sets of pitches and organizing music around those sets dates back hundreds of years. A Pitch Class Set is simply an unordered collection of pitches. There are 12 unique pitches or pitch classes on a keyboard, and these are numbered from 0 to 11, starting with 'C'. For example, the pitch class set consisting of the notes C, E, and G would be written as (0,4,7).

A set can be defined by any collection of objects. From the world of science, we have an eclectic slew of sets you are already familiar with. We have sets of planets, mountains, rivers, fish, amphibians, reptiles, mammals, plants, grasses, flowers, trees, and on and on. From the world of medicine, we can identify sets of prescriptions, diseases, patients in a hospital on a given day, known cures, known illnesses, appendages, medicines, and bandages, just to name a few. There are sets of colors, sets of months, sets of staplers sold at Office Depot on June 12, 2004, sets of monthly bills, sets of famous people, sets of football players, sets of Enron executives, sets of email accounts on Yahoo, sets of insurance companies, sets of songs that rank Number One on Billboard’s list, sets of serial killers, sets of vice presidents, sets of best friends, sets of baseballs hit out of Wrigley Field. Any unordered collection of objects will suffice to form a set. Those individual pickles consumed in Maryland last evening form a set.

World of Numbers Mathematicians use Set Theory to classify numbers into many different types of sets. Some of the more common set types include counting numbers, integers, rational numbers and real numbers. The counting number, otherwise known as natural numbers are defined at the set  = {1, 2, 3, 4, ... }. The integers, all the counting numbers, zero and all of the negative whole numbers, are traditionally defined with a Z. The letter Z is taken from the German word Zentrum, which means integer. Any number that may be written as a fraction is called a rational number. A rational number is simply a ratio. Traditionally, this set is denoted with the capital letter Q which comes from the word quotient. The set of real numbers can be visualized with the number line, extending from negative infinity to positive infinity. Each and every point along this line represents a real number.

Lastly, if we wanted to enhance our visualization of two arbitrary two sets, set A and set B, we could see them as over lapping figures, potentially sharing some common ground, with the common ground referred to as the intersection of the two sets. We shaded each respective set, and in the first problem, we will include visual cues to help us see the regions described.

20 U Set A 

U Set B 

Problem One

Let U = {x | x is a student at American university}. A = {x | x is a full-time student at American university}. B = {x | x is an American university student who works 20 or more hours each week}. Find and describe the following: a. A\B b. B\A c. A  B d. A ∩B e. A f. B g. A  B h. A ∩B

Solution a. A\B = {x | x is a full-time American University student who works less than 20 hours each week}. U Set A\B 

21 b. B\A = {x | x is an American University student who works more than 20 hours each week but is not full time} or {x | x is a part time American University student who works more than 20 hours each week} U Set  /  

c. A  B = {x | x is a full time American University student or one who works more than 20 hours each week}

Set    U 

d. A ∩B = {x | x is a full time American University student who works more than 20 hours each week}

Set    U 

e. A = {x | x is a part time student at American University} U Set  

22 f. B = {x | x is a student at American University who does not work 20 hours per week}

U Set  

g. A  B = {x | x is not a full time American University student and does not work more than 20 hours each week} U Set    

h. A ∩B = {x | x is all students at American University who are not full time working 20 hours per week}

U Set    

Problem Two

U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1,2, 3, 4}, B = {1, 2, 6,7}, C = {2, 3, 6, 8}. Find the following: a) AU B b) B∩ C c) A∩ C d) A∩ B ∩ C e) C \ A

23 Solution a) AU B consists of all of the elements in both set A and set B. AU B is {1, 2, 3, 4, 6, 7}. b) B∩ C consists of all of the elements that are in both sets B and C, the overlap of B and C. B∩ C = {2, 6}. c) A∩ C consists of the overlap of A ={1, 2, 3, 4} and those elements that are not in C = {0, 1, 4, 5, 7, 9}. A∩ C = {1, 4} d) To find the elements in the set A∩ B ∩ C , we first find the elements that are not in the overlap of sets B and C. B∩ C ={2,6} = {0,1,3,4,5,7,8,9} . So, now we find the elements that are in the overlap of {1,2, 3, 4} and {0, 1, 3, 4, 5, 7, 8 9}, which is (1, 3, 4} e) C \ A are the elements in set C after we remove any elements that are also in set A. So, we start with {2, 3, 6, 8} and we remove the elements that are in set A too, so we remove {2, 3}, leaving {6, 8}.

Problem Three

Let U = {x|x is a student at Washington Middle School, grades seven to nine}, A = {x|x is a seventh grade student at Washington Middle School }, B = {x|x is a student at Washington Middle School enrolled in art}. Find and describe the following: a) A∩ B and b) AU B

Solution a) A∩ B are those students at Washington Middle School who are not the seventh graders taking art. This would be that same as those eighth and ninth graders or any student not taking art at Washington Middle School. b) AU B are those students at Washington Middle School who are not the seventh graders or not taking art. This would be that same as those eighth and ninth graders that are not taking art at Washington Middle School.

In our concise notation for set theory, we call these laws that are true for any two sets, DeMorgan’s Laws: A∩ B= AU B and AU B= A∩ B

Problem four

Find the following sets: a)  \  b)  \ c)  \ Q

Solution a)  \  are the negative integers as well as 0. b)  \ are the counting numbers that are not real numbers. Since all counting numbers may be found on the real number line, this set is empty, denoted .

24 c)  \ Q are the real numbers that may not be written as a fraction. These numbers have a name, they are called the irrational numbers and include numbers that have decimal expansions that do not terminate or follow a pattern, like p , 2 or e.

Problem Five

U = {x|x is a card in a standard deck of 52 cards}, A = {x|x is a two}, B = {x|x is a black card}, C = {x|x is a jack}. Find the following: a ) n( AU B ) b) n( B∩ C ) c) n( CU ( B \ A ) Solution

We are looking for the number of elements in each set. So, we will find the elements in each set and count the number of elements. a) AU B consists of all two’s and all jacks in a standard deck of cards. So, there are 4 of each (2 of hearts, 2 of diamonds, 2 of clubs, and 2 of spades as well as the jack of hearts, the jack of diamonds, the jack of clubs and the jack of spades. So, n( AU B ) = 8. b) B∩ C consists of the black jack, of which there are two of them, the jack of clubs and the jack of spades. c) CU ( B \ A ) consists of all of the jacks as well as the cards in the set B\ A . The cards in the set B\A are those black cards that are not two’s. So, B\ A is the set of all of the other cards, of which there are 26 red cards, and the 2 black two’s. This set of 28 cards consists of two of the jacks. So, we must add in the other two jacks. So, n( CU ( B \ A ) = 28 + 2 = 30.

Exercise Set

The set denoted U is the Universal set for C= {x|x is a person who does not own a that problem. For numbers 1 – 16, cell phone}, describe the following sets in a complete 4. A∩ C sentence. 5. A∩ B

n( A∩ B ) Let U = {x|x is a US citizen in 2004}, 6. A = {x|x is a person who has purchased from Amazon.com}, Let U = {x|x is a perspective home buyer B = {x|x is a person who does not own a in your city}, cell phone}. A = {x|x is a person could qualify for up to a $ 250,000 home}, 1. A∩ B B = {x|x is a person who is ready to 2. A \ B purchase a home today}, and 3. AU B C = {x|x is a person who wants to list their house for sale}. Let A = {x|x is one of the original signers 7. AU( B∩ C ) of the Declaration of Independence}, 8. A∩( BU C ) B = {x|x was a president of the US}, 9. A∩( B ∩ C )

25 Let U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, Let U = {x|x is a registered voter}, A = {0, 2, 4, 6, 8}, B = {1, 2, 3}, A = {x|x is a person who is opposed to the C = {2, 3, 4, 5, 6}. death penalty}, Find the following: B = {x|x is a person who is opposed to 17. AU B legalized abortions for any reason}, and 18. B∩ C C = {x|x is a person who is a registered 19. A∩ B democrat}. 10. C \ A 20. AU C∩ B 11. A∩( BU C ) 21. A \ C 12. C∩( B ∩ A ) Let U = {x|x is a card in a standard deck of 52 cards}, A = {x|x is a black You have a job where you need to come card}, up with a data bank of known suspected B = {x|x is a jack}, C = {x|x is a diamond}. terrorists. The government is looking into Find the following: links between terror cells in various A B countries, trying to find patterns of 22. U cooperation. You are investigating cells in 23. B∩ C countries who are not known to have links. 24. AU C 25. B \ C Let Let U = {x|x is a card in a standard deck of U = {x|x is a known suspected terrorist}, 52 cards}, A = {x|x is a heart}, A = {x|x is a known suspected terrorist B = {x|x is an ace}, who once lived in Saudi Arabia}, C = {x|x is a black card}. B = {x|x is a known suspected terrorist Find the following: who once lived in Chechnya}. 26. n( AU C ) Find and describe the following: 27. n( B∩ C ) 13. A∩ B 28. n( BU A ) 14. AU B 29. n( C \ ( BU A ) Let U = {x|x believes in mysticism and the Let U = {x|x is a card in a standard deck of occult}, A = {x|x believes in Tarrot cards}, 52 cards}, A = {x|x is a B = {x|x believes in channeling mediums} diamond}, and C = {x|x believes in crystal powers}. B = {x|x is a six}, Find and describe the following: C = {x|x is a face card (Jack, Queen or 15. (C∩ A ) \ B King)}. 16. C \ A Find the following: 30. n( AU C ) 31. n( B∩ C ) 32. n( A∩ ( BU C )) 33. n(( B∩ A ) \ C )

26 B = {x|x is a refrigerator on sale at Sears}, C = {x|x is a microwave on sale at Sears}. Let A = {x|x is a resident of the state of Find and describe the following: Arizona}, B = {x|x us a college student}, 43. BU C C = {x|x is a smoker}. 44. A∩ C Find and describe the following: 45. AU C 34. AU C B∩ C 35. Let U = {x|x is doctor}, 36. A∩ C A = {x|x earns over $ 200,000 per year}, 37. A \ C B = {x|x is a male}, 38. A∩( C \ B ) C = {x|x is not taking any new patients}. Find and describe the following: Let U = {x|x is a letter of the alphabet}, 46. B∩ C A = {x|x is a vowel (a, e, i o, u)} 47. AU C B = {x|x is a letter that comes before g}, 48. B\ C C = {x|x is p or u}. Find the following: Let U = {x|x is a card in the standard deck 39. n( A∩ C ) of 52 cards}, A = {x|x is a red card}, 40. n( B∩ C ) B = {x|x is a face card (jack, queen or kind)}, C = {x|x is a back jack, black 41. n( A∩ BU C ) queen, black kind or black ace}. n(( A∩ B ) \ C ) 42. Find and describe the following: 49. n(( BU C ) \ ( A∩ B )) Let 50. n( C∩ ( BU A )) U = {x|x is an appliance for sale at Sears}, A = {x|x is an dryer on sale at Sears},

27 Multiplication Principle

Each morning, each of us basically has the same routine. After awaking, showering, and brushing our teeth, we then must decide what to wear for the day. Suppose, for the sake of discussion, we are open our closet and find there are four shirts, one tie-died, one blue, one green and one yellow. We have two jeans, one black, one blue, and three pairs of foot wear, one pair of sneakers, one pair of sandals and one pair of loafers. How many outfits can we wear that day? One. How many possible choices do we have for that one out fit?

Well, for starters, with each shirt, we can pair it up with any one of two jeans. We can pair the tie-died shirt with the black or blue jeans (two choices), the blue shirt with the black or blue jeans (two more choices) and the same two choices for pairing each of the green and yellow shirts (two choices a piece.) In other words, each time we pair four items of clothes with two other items of clothes, we are saying four groups of two, which is the repeated addition of 2+2+2+2 or more easily, 2 times 4 = 8 choices. This is a basic definition of multiplication.

So, extending this thought, we have 4 shirts that can be paired with any one of two pairs of pants and then each pairing of shirt and pants can be then paired with any one of three pairs of shoes. In other words, we have 4 * 2 * 3 choices, or 24 choices for outfits before us. Now, this has nothing to do with common sense, right? I mean, who would wear a yellow shirt, black pair of pants and loafers?

The same principle may be applied when an eager child visits Baskin and Robbins Ice Cream Shop, better known as 31 Flavors. The child looks to order a triple scoop ice cream cone. Even if the child doesn’t want a flavor to repeat, the he or she has to peak into 31 different tubs of ice cream to decide among 31 x 30 x 29 or 26,970 different possible triple scoop cones. No wonder kids takes so long ordering.

People don’t realize the sheer power of numbers, and as a result, you hear the darndest things. Here in Phoenix, the Suns had been to the NBA Finals in the 1990’s, and a few years back, around the turn of the century, the Diamondbacks actually won the world series. But, recently, these sports teams have fell on difficult times. This story is absurd, but true. Last week, we were lucky enough to attend both a Suns game and a Diamondbacks game the same week. Unfortunately, we lost both games. Whenever one goes to these games, there is always a fan, who clearly disgruntled by the teams futility, shouts something like, “We got the talent, just find the right combination of five that will gel.” “You got the players, just try every line-up until you find the one that works.”

Of course, upon hearing this, placing down our beverages, ignoring the truism that ‘silence is golden’, we always saunter up to the misguided soul with a swiftness that usually stuns him. “Let’s examine that,” we begin, as the distracted fan tries to bend his head around us to see the game. Oblivious, we continue, and at the Suns game, it went like this. “There are 12 players on the team. That’s 12x11x10x9x8 possible combinations or … “ and take out the trusty pocket calculator because we are math nerds, and conclude with “95,040 permutations we must endure until we find the right one. Even if each set of five were on the court for just

28 one minute, that’s 95,040 divided by 48 minutes per game or 1,980 games we would have to play and at 82 games a season, that’s over 24 seasons!” We leave the fan confused.

Of course with baseball, the necessary modification to the argument made the numbers even more preposterous. It went some thing like this “25x24x23x22x21x20x19x18x17 or 7.41354768 x 1011 line-ups”, and we would end the conversation with “the earth is not going live long enough to fulfill your asinine request.”

Notation: 12 factorial = 12! = 12x11x10x9x8x7x6x5x4x3x2x1 = 479,001,600 There are more typical examples of the multiplication principle, or continuous branching, that we all encounter every day. You’re in LA, you need to fly home to NYC. The airlines tell you that you can fly to Chicago, Milwaukee, Denver or Atlanta. From each of these cities, you can fly to the New York area and arrive in either Kennedy, Laguadia or Newark. This branching is quite visual, you can almost see the number of choices before your eyes, each of the four Midwest cities is paired with the three cities in the NYC area, so you have four times three, or twelve different flight schedules to choose from.

Credit card numbers vary from 15 to 20 digits long. If each digit could be any one of ten numbers, 0, 1, 2, 3, … , 9, there are 10x10x10x … x10 or 1015 to 1020 possible credit card numbers. That’s somewhere between 1,000,000,000,000,000 or 100,000,000,000,000,000,000 credit card numbers. And since there are 6.3 billion people on this planet, 1020 divided by 6.3 billion is roughly 1.6 x 1010 or nearly 16 billion times the population of the world. If all credit card numbers were 20 digits in length, each human on the planet would be able to choose from 16 billion different numbers. What are the odds that a would be thief would find your credit card number? Oops, that’s the next chapter, but it is quite unlikely, wouldn’t you think?

We all have encountered the need to create a Personnel Identification code. Maybe it was done online (email accounts, bank accounts, stock market inquiries, club member, to secure airline tickets) or with an ATM machine. Now, we ask two fundamentally different questions. Please think about the answer. If the Personnel ID code could consist of four numbers, how many codes are possible? If the Personnel ID code could consist of 4 to 8 numbers, how many possible ID codes are there? Again, these are two very different questions. Do you see the difference?

For the four number code, there are four slots, and each slot may have any one of the ten digits filling the slot. So, each of the ten digits in the first slot is paired with each of the ten digits in the second slot and each of these (10 x 10) pairs are paired with any of ten digits in the third slot and so on. There are 104 or 10,000 possible codes.

If we had a choice of Personnel ID codes, where we may have anywhere from 4 to 8 digits, then we may have 104 or 105 or 106 or107 or108 choices, depending upon if we want a code with 4, 5, 6, 7 or 8 digits, respectively. Since we will have either a 4 digit code (10,000) or a five digit code (100,000) or a six digit code (1,000,000) or a seven or eight digit code (10,000,000 or 100,000,000) we would add up these different possibilities. There are 111,111,000 codes possible if we can have a choice for our code to be 4 to 8 digit in length.

29 Again, this underlying principle is called the multiplication principle (aptly named, don’t you think?) But, in the branch of mathematics called combinatorial mathematics, we find there can be different versions of the question “in how many ways can something be done?” and each version represents a different scenario.

Permutations and Combinations.

Suppose your school mathematics club has thirty students, and you need to form a group of three from the set of 30 students. How many possible ways can we do this?

The underlying crucial question needed to be asked is what is the purpose of this group. In combinatorial mathematics, we use the phrase does order matter? What do we mean by order? To answer this question, we must ask ourselves what are the groups for because the roles the people take on within the groups can be categorized into two separate and very different scenarios, scenarios which are as different as night and day.

Scenario 1: Permutations. In this scenario, the three people in the group are going to serve political offices, the group will have an elected president, vice president and treasurer for this club. Since each of the three people chosen serve different roles, different political offices, we use the term order to refer to these differences. In other words, from the set of 30 students, let’s examine just one group of three, Tommy, Malcolm and Vanessa. For these three people, it matters greatly what role they serve, ie, it matters how they are ordered. It matters to Vanessa whether she is the president, vice-president or treasurer. The same may be said for Tommy and Malcolm. Each of the three people serve different roles.

Scenario 2: Combinations. In this scenario, the members from the group are going to serve on a committee of three for say a campus beautification project. Each of the three people chosen serve the exact same role on the committee. They are interchangeable. Thus, the order they are chosen doesn’t matter, since they are all committee members. Tommy, Malcolm and Vanessa do not care whether they were chosen first, second or third because they will all serve equally on the committee.

The difference between a permutation and a combination is the notion of order. The numbers of permutations refers to the number of possible ways you may arrange elements from a set where order matters. Compiling lists for political offices, different line-ups for sports or debate teams, investments where each has a different risk, all implicitly convey a sense of order. If you were selected to be a member of one of these groups, it would matter which role you served. The number of combinations refer to the number of possible ways you may perform a task where order does not matter. Members on a committee, number of ways to arrange colored lights, investments where each has the same risk, imply order doesn’t matter. If you were a member in any of these groups, it would not matter which member you were or the order in which you were selected.

Permutations. From a club of 30 people, how many possible ways can we elect a president, vice-president and a treasurer? Well, let us begin with the multiplication principle. Any one

30 of the 30 people can be the president, once the president is chosen, then any one of remaining 29 people could be the vice-president, and then anyone f the remaining 28 people could be the treasurer. Said another way, any one of 30 people could be paired with any one of 29 people, 30 x 29 = 870 and any one of the 870 couplings can be paired with 28 people, 870 x 28 = 24,360 different groups of three. Interpretation, from a club of 30 members they are 24,360 ways to elect a president, vice-president and a treasurer. Thus, we say there are 24,360 permutations to fill three offices from a group of 30.

Notation: 30P 3 = P(30,3) = 24,360 P= P( n , r ) = n ( n - 1)( n - 2)...( n - r + 1) General Permutation Notation: n r 1 4 4 4 4 2 4 4 4 4 3 r factors

The number of permutations may also be thought of as the arrangements of r items selected without replacement from a pool of n items. Permutations are used whenever order n! is important and an alternate formula is given by P= P( n , r ) = . n r (n- r )!

Combinations. From a club of 30 people, how many possible ways can we select an ad-hoc committee of three? Again, let us begin with the multiplication principle. Any one of the 30 people can be selected first, then any one of remaining 29 people could be selected next, and then anyone of the remaining 28 people could be selected last. So, 30 x 29 x 28 = 24,360 different groups of three could be chosen. But, here lies the rub. For each group of three, the order doesn’t matter since the three people are all serving the same role, holding the same office. Now, we ask ourselves, how do we get rid of the duplicates because order does not matter? Well, first, in how many ways can we order a group of three? 3 x 2 x 1 = 6. So, if we divide out the number of ways each group of 3 can be ordered, we have the total number of possible groups of 3 formed from a pool of 30 people; 24,360/6 = 4,060. Thus, there are 4,060 ways to elect a committee of three from a group of 30 people, or 4,060 combinations of three people to fill committees of size three from a group of 30. When we use combinations, we tend to use the vocabulary ‘choose’ to indicate the fact that order doesn’t matter. We say 30 choose 3 to mean C(30,3). It is more succinct than saying “in a group of 30 people, we need to form a group of three where order doesn’t matter.” To say 30 choose 3 is colloquial but that is ok, it is accepted mathematical terminology probably because it is so short and so precise.

P(30,3) 30 x 29 x 28 Notation: 30C 3 = C(30,3) = = = 4,060 3! 3x 2 x 1

n( n- 1)( n - 2)...( n - r + 1) P( n , r ) 1 4 4 4 42 4 4 4 43 General Combination Notation: C= C( n , r ) = = r factors n r r! r ( r- 1)( r - 2)...2 x 1

31 The number of distinct combinations of r items selected without replacement, when the order of the selection is not important, from a pool of n items may also be designated with an n! alternate formula is given by C= C( n , r ) = . n r r!( n- r )!

Next we will interchange the fundamental multiplication principle with permutations and combinations and decipher which is which.

Problem One

A town meeting of local political leaders consists of 8 democrats, 6 republicans and 4 independents. Suppose the group wants to elect each of the following. How many ways are there to do this? [Hint: The last two parts, g and h, are remarkably different than the first 6 parts. Why?] a) a president, a vice president and a treasurer b) a president who must be a democrat, a vice-president who must be a republican and a treasurer c) a committee of five, where the chair is a democrat d) a committee of five, with two democrats e) a committee of five, the chair is republican, and the rest are independents f) a committee of five, where the chair is not a democrat, but the remainder of the committee is democrats g) a committee of five with at least two democrats h) a committee of five with at most one democrat

Solution

Our problem spans all three combinatoric principles. a) There are 8+6+4 total people to choose from for president. Since a person can not hold more than one office, once the president is chosen, there is one less person that could be the vice president. So, there are 18 choices for president, 17 choices for vice president and 16 choices for treasurer. There are 18(17)(16) = 4896 choices to fill the three offices. b) There are 8 democrats who could be president, 6 republicans to choose from to be vice president and 4 independents who could be treasurer. There are 8(6)(4) = 192 possible ways to fill these offices. c) There are 8 ways to fill the chair, any of the 8 democrats could be chair. Then we must fill the committee of the remaining four from the other 17 people left over. 骣P(17,4) 17(16)(15)(14) There are 8*C (17,4)= 8琪 = 8 = 17(16)(5)(14) = 19,040 ways. 桫 4! 4(3)(2)(1) d) We need to find how many combinations of democrats, this would be choosing 2 from 8 people or 8 choose 2 or C(8,2). We need to pair these (multiply) with the remaining 3 people that we choose from the remaining 10 leaders, or 10 choose 3, C(10,3). So, C(8,2) = 8(7)/2 = 28 and we need to pair the 28 combination of democrats with each of the C(10,3) = 10(9)(8)/3! = 90(8)/6 = 120 possblle combinations of 3 from 10 people. Since we are forming the whole committee of size

32 5 (remember these words) we multiply. There are C(8,2)*C(10,3) = 28*120 = 3,360 ways to form these committees. e) Ok, we need a committee of five, and for the chair is republican, there are 6 choices. Thus, we must choose the remaining four from independents, so that is C(4,4) = P(4,4)/4! = 4*3*2*1/(4*3*2*1) = 1. So, we have 6*1 = 6 possible choices. f) For a committee of five, where the chair is not a democrat, there are 10 choices (6 republicans plus 4 independents) and since remainder of the committee are all democrats, we multiply by 8 choose 4. 骣P(10,4) 10(10*9*8*7) 10C (10,4)= 10琪 = = 2100 ways. 桫 4! 4*3*2*1 g) This problem is different than the others we faced because there are many ways to form this committee of five with at least two democrats, we could form a committee with 2 democrats, 3 democrats or 4 democrats or 5 democrats. We must add the number of ways to form a committee of 2 democrats to the numbers of ways to form a committee with 3 democrats because we would do one or the other but not both. We then add the number of ways to form 4 and 5 democrats respectively. C(8,2)* C (10,3)+ C (8,3)* C (10,2) + C (8,4)* C (10,1) + C (8,5)* C (10,0) = 3360+ 2520 + 700 + 56 = 6636 h) Again, there is more than one way to form a committee of five with at most 1 democrat. We could form a committee with no democrats, 1 democrat or 2 democrats. C(8,1)* C (10,4)+ C (8,0) * C (10,5) = 1680+ 252 = 1932

Problem Two

From a standard deck of 52 cards, how many 5 card hands are possible … a) that have four two’s b) that have at least two two’s. c) a royal flush (10, J, Q, K, A of the same suite.) d) a hand without a royal flush

Solution

For poker hands, the key question is “does the order matter?” because this will determine whether we are dealing with a permutation or a combination. By order, we can visualize the order of the cards as they arrive into our hand from the deck. If you were dealt four two’s and a seven, would it matter whether the seven came first or second (or third or fourth or fifth?) No, of course not. So, we are dealing with combinations and we will use the verbiage choose to help us along. a) So, for four two’s to be dealt, the firth card must be a non-two. We are forming the whole hand, so we have 4 choose 4 (4 two’s from 4 two’s) and the remaining 48 cards, we choose 1. So, C(4,4)* C (48,1)= 1(48) = 48. 48 ways to get 4 two’s in a 5 card hand should make perfect sense, since there is only one way to choose

33 4 two’s from 4 two’s and this combination is paired with any of the remaining 48 cards. b) To be dealt at least two two’s, there are 3 different ways to do this, you may be dealt 2 two’s, 3 two’s or 4 two’s. One hand is not all of these possibilities, rather are either dealt 2 two’s or 3 two’s or 4 two’s. So, we add the number of ways to get 2 two’s to the number of ways to get 3 two’s to the number of ways to get 4 two’s. C(4,2)* C (48,3)+ C (4,3)* C (48,2) + C (4,4)* C (48,1) = 1 4 4 2 4 4 3 1 4 4 2 4 4 3 1 4 4 2 4 4 3 2two ' s 3 two ' s 4 two ' s 48+ 4512 + 103776 = 108,336 ways c) common sense tells us there should be 4 ways to be dealt a royal flush (10, J, Q, K, A of the same suite), one way for each suite. How do we calculate this? 4*C (5,5) C (47,0) 1 442 4 43 = 4(1) = 4 this is the calculation for each of the4 suites d) To calculate all of the possible hands that are not royal flushes directly, it would take us a long time because we would have to consider all of the possible hands that are not royal flushes. So, instead, we will do this calculation indirectly by subtracting the ways of a small known quantity. We will subtract the 4 ways to get a royal flush from all of the possible hands that could be dealt with five cards. This number is just 52 choose 5, right? So, C(52,5)= 2,598,960 five card hands possible. Thus, there are 2,598,960 – 4 = 2,598,956 possible hands that are not royal flushes.

One last problem that contains a review of the  multiplication principle  where there are several different  permutations - order matters ways to form the specified group  combinations - order doesn’t  the indirect method where doing matter the problem directly would take  both the multiplication principle too long and combinations

Problem Three

Suppose, there are 4 men and 8 women. We want to choose a committee of size 4. In how many ways can we … a) select a man as the chair, a woman as alternate chair, a man as the treasurer and a woman as the public speaker. b) select an all woman committee, where the four positions to be filled are chair, alternate chair, treasurer and the public speaker. c) select the committee of four. d) select a man as the chair, the remaining committee members are women e) select at least three men on the committee. f) not select a committee with all men.

Solution

34 a) 4(8)(3)(7) = 672 ways taken from the multiplication principle. b) P(8,4) = 8(7)(6)(5) = 1680 permutations for this to occur. c) C(12,4) = 495 ways to choose groups of 4 from 12. d) 4*C(8,3) = 224 ways to form this committee.

e) There are two possible types of committees, one that has 3 men and one that has 4 men. We must add the number of 3 men committees to the number of 4 men committees. C(4,3)*C(8,1) + C(4,4)*C(8,0) = 32 + 1 = 33 ways. f) There are too many possibilities to consider, one man, two men, etc . But, if we have 4 men and 4 slots to fill on the committee, there is only 1 way to fill a committee with all men. Thus, there must be C(12,4) – 1 = 495 – 1 = 494 ways to fill a committee not with all men.

Problem Four

You are visiting a nice family for the holidays. There house is lovely and it is Christmas eve. Everyone is busily decorating. In a corner by the tree is a box of lose Christmas lights. In the box are precisely four green Christmas lights, five white Christmas lights and eleven red Christmas lights. You ask, “hey, what should I do with this box?” The hostess replies, “Just string the lights together, any way you want.” You hands start to tremble. You feel sweaty. You know you are terribly indecisive, and you fear you may obsess over this seemingly trivial task. Are you justified in your worry and by the way, how many possible arrangements are there for you to obsess over if you were told to string together the lights?

Solution

In any arrangement, any two lights of the same color are interchangeable. They are like committee members, they fulfill the same role. Thus, this calculation works similar to a committee, we need to divide out the number of possible ways to order the same color. That said, this is the multiplication principle, the number of ways to arrange 4 + 5 + 11 or 20 is 20!= 2.43x 1018 arrangements for 20 items. But since we are only dealing with three interchangeable colors, we may divide out the number of ways to order 4, 5 and 11 items, respectively. Still, if you are of the obsessive nature, you have to sift through 20! = 21,162,960150 arrangements before settling on the one you like best. I hope its 4!5!11! early in the evening on Christmas eve and you work really fast.

Problem Five

Try this the next time you are in a crowd. Like a family get together, a party or the next time you go to class. Gather everyone and have them stand in a circle. Count the number of people in the circle. Then have everyone shake hands with each other once. I have two questions. How many hands were shook and how many handshakes occurred?

Solution

35 How many hands did each person shake? For example, if there were 19 people, then each shook 18 hands. How many hands were shook? Well, if there were 19 people, each person shook 18 hands, so, 342 hands were shook because 19(18)  19 P2 How many handshakes occurred? Well, if there were 19 people and each person shook 18 hands then for each of the 342 hands that were shook, each a handshake was counted twice, so there were actually ½ of the 342 handshakes, or 171 handshakes. In other words, 342 19(18) handshakes where we divide out the order, 2. We have  C handshakes. 2*1 19 2

One last note about ordering items. In case you are wondering, factorials grow fast. Real fast. Most calculators, like the TI 82, TI 83 can only handle 69!, which is 69!= 1.71x 1098 . 70! causes an overflow because the number is too large. This means something to us. Ordering 69 objects, P(69,69), well there are more ways to do this than there are molecules in the universe. Be careful with errant statements with “let’s try all possible orders before we decide.” Sloppy words, sloppy thought processes. And this is exactly what we are trying to avoid. Next time someone tells you to try all flight schedules, try all batting line-ups, string together the lights any way you want, remember this moment.

Exercise Set

1. How many license plates consisting of 3 c) Select a committee of five with 2 letters followed by 4 digits are possible if freshman. a) repetitions are allowed? d) Select a committee of five with at least b) repetitions are not allowed? 4 freshmen. c) no even digits or repetition of digits are e) Select a committee of five with at most allowed? 4 freshman. d) An X or a Q must be the first letter? f) Select a committee of five with no e) The four digit number is even? freshman. f) The four digit number is less than g) Select a committee of five where the 1000? chair is a senior. g) Order, from the most number of license h) Select a committee of five where the plates possible to the least number of chair is either a junior or a senior. license plates possible, a) through f) i) Select a committee of five where the above. Does the order match your chair is not a senior. intuition as you read the specifics of how these license plates were 3. I open my closet and see 4 pants, 7 produced? shirts, 3 ties and two pairs of shoes. How many outfits are possible? 2. In a group of 5 freshman, 9 sophomores, 6 juniors, and 8 seniors, how 4. How many ways are there to arrange 5 many ways can we … people in a line? a) Elect a president, vice president and treasurer. 5. An investor is going to invest $ 50,000 b) Select a committee of six consisting of in 4 stocks from a list of 10 prepared by exactly 2 seniors and exactly 3 juniors.

36 his broker. How many different d) how many hands with three threes are investments are possible if possible? a) $ 12,500 are invested in each stock. e) how many hands with three kings and b) $ 10,000 is invested in three of the two aces are possible? stocks, $20,000 is invested in the last f) how many full houses, three of a kind stock. and two of a kind, are possible?

6. There are ten men and twelve women in a club. 11. If we rent 5 videos from a store, one a) In how many ways can a committee of comedy, one action, one mystery, and two four be formed? documentaries. b) In how many ways can a committee of a) how many orders can we view all 5 four be formed, if a woman must be the videos, seeing each one just once? chair? b) how many orders can we view 3 c) In how many ways can a committee of videos, seeing each one just once? four be formed if they are all women? c) how many orders can we view 3 d) In how many ways can a committee of videos, seeing the comedy last? four be formed if there are at least 1 d) how many orders can we view 3 man? videos, seeing a documentary first and e) In how many ways can a committee of last? four be formed if there are at most 2 e) how many orders can we view 3 woman? videos, seeing both documentaries?

7. How many different 8-letter words (real 12. How many ways are there to arrange 4 or imaginary) can be formed from the paperbacks and 5 hard cover books on a letters in the word APPROACH? shelf, if the paperback books are in a row and the hard cover are in a row? 8. How many ways can we arrange 8 roses, 4 violets and 5 daises in a row along a 13. Tests border of a garden? a) How many ways are there to answer a 20 question true-false exam? 9. There are 125 people and three door b) How many ways are there to answer a prizes. 20 question multiple choice exam if a) How many way can 3 door prizes of $ there are 4 choices per question. 50 each be distributed? c) How many ways are there to answer a b) How many ways can door prizes of 20 exam where 15 questions are true- $5000, $ 500 and $ 50 be distributed? false and 5 questions are multiple choice with 4 choices each? 10. From a standard 52 card deck, if we were playing poker where 5 cards are dealt 14. Combination Locks to each player, a) A lock has 50 digits, and the a) how many unique 5 card hands are combination involves turning right possible? to the first numbers, turning left to b) how many flushes are possible? the second number and turning c) how many one pair hands are possible? right to the third number. How

37 many possible combinations are 19. There are 7 boys and 5 girls on a co- there? ed basketball team. A team plays 5 b) A lock has 50 digits, and the players on the court at one time. combination involves turning either a) How many possible ways are their to left or right to the first numbers, field a team of 3 boys and 2 girls? turning either left or right to the Position doesn’t matter. second number and turning either b) We can play at most 3 boys. Position left or right to the third number. doesn’t matter. How many possible combinations are there? 20. You must take math, science, business and one elective all on a MWF class 15. There are 6 speakers at a conference, schedule. Ms. Ali, Ms. Boo-Koo, Ms. Campbell, Mr. a) how many class schedules are Doyle, Mr. Evy, and Mr. Fritch. possible? a) In how many ways can we order the b) how many class schedules are possible speakers so that Mr. Fritch speaks if the math must be taken before the before Mr. Evy? science? b) In how many ways can we order the c) how many class schedules are possible speakers so that the women speak last? if the math class can not be the first c) In how many ways can we order the class of the day? speakers so that Mr. Fritch doesn’t speak first? 21. How many possible 5 card poker hands can be dealt: 16. In how many ways can three different a) if we knew all of the cards in our hand plants be line up on a window sill if we are were spades choosing from 6 plants? b) if we knew three of the cards in our hand were spades 17. If Frosty Ice Cream has 40 flavors, c) if we knew there were no more than a) how many possible 2 scoop cones are three of the cards in our hand that are possible? spades b) how many possible 3 scoop cones are d) if we knew there were three aces in our possible? hand c) how many possible 2 scoop cones are e) if we knew there were no aces in our possible, if chocolate is the bottom hand scoop? d) how many possible 2 scoop cones are 22. Below are the choices on has when possible if chocolate is not the bottom they purchase Brand X car. scoop? Color Red Yellow Blue e) how many possible 2 scoop cones are Model Deluxe Standard possible if the scoops must be different? Engine Standard Automatic How many choices does the consumer 18. How many ways are there to arrange a have in their purchase? string of Christmas lights if there are 20 green bulbs, 10 white bulbs and 15 red 23. How many possible 5 card poker bulbs? hands can be dealt: a) containing three sixes b) containing four sixes

38 c) containing four sixes or four fives d) containing four of a kind e) containing at least three sixes f) containing three sixes and two fives 26. How many passwords are possible if g) containing a flush, all hearts they comprise of … h) containing a flush, any suit a) 4 digits b) 4 – 8 digits 24. How many winning lottery numbers are possible of a pick 5 where you must 27. How many passwords are possible on match all five numbers from 50 balls. a ‘hotmail account’ if the password consists of six-character minimum, 25. Which lottery would be easier to win, twelve-character maximum? No spaces, picking 5 correct out of 50 or 6 correct out but the password can include digits, letters of 60? and the underscore. Assume CAPS or small case yields the same password.

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