Math 510 Fall 2008 Exam 1 Solutions October 3, 2008 ———————————————————————————————————————–

1. (a) Suppose that the first 20 kids into Fenway get one of 20 bobbleheads, each of a different player. How many different ways are there to pass out the 20 bobbleheads? We are just asking for permutations of a set with 20 objects. This is P (20, 20) = 20!.

(b) Suppose instead of 20 different players on the bobbleheads, there are only 4 distinct bobblehead types for each of the 4 Red Sox who started the All-Star game. There are 6 bobbleheads, 4 of , 2 of , and 8 bobblheads. Now how many different ways are there to pass out the 20 bobbleheads? This question is asking for permutations of a multiset with k = 4 types of objects n1 = 6 of the first kind, n2 = 4 of the second kind, n3 = 2 of the third kind, and n4 = 8 of the fourth kind. The formula tells us the number of perumtations of this set of size 20 is n! 20! = . n1!n2!n3!n4! 6!4!2!8!

2. Perhaps the Red Sox mascot, Wally the Green Monster, has a sweet tooth. So the players get together to buy him 6 gourmet chocolates from a candy store. If the store stocks 10 different kinds of candy, how many ways can the players select a bag of six pieces of candy, each piece of a different kind? 10 We are asking for 6-combinations of 10 pieces of candy. That is just 6 .

3. Determine the number of positive integers that are factors of the number

35 × 73 × 396 × 53.

Any factor will look like 3i × 7j × 39k × 50` where i is between 0 and 5, j is between 0 and 3, k is between 0 and 6, and ` is either 0 or 1. This gives 6 options for i, 4 options for j, 7 options for k, and 2 options for `. Using the multiplication principle we have 6 × 4 × 7 × 2 = 336 factors.

1 4. How many integral solutions of

x1 + x2 + x3 + x4 + x5 = 29

satisfy x1 ≥ 0, x2 ≥ 3, x3 ≥ −2, x4 ≥ 7, and x5 ≥ −1?

We substitute variables y1 = x1, y2 = x2 − 3, y3 = x3 + 2, y4 = x4 − 7 and y5 = x5 + 1. So we are asking equivalently for positive integral solutions to y1 + y2 + y3 + y4 + y5 = 29 − 3 + 2 − 7 + 1 = 22. This is just the number of permutations of a multiset with 22 26! of one object and 4 of another object which is 22!4! .

5. (a) Assume the Red Sox and their staff use three buses to get all 48 of them from Fenway to the airport for a road trip. If each bus seats 16 people and the 6 coaches all are to be on the same bus, how many ways are there to distribute everyone among the three buses. (Where each player sits on an individual bus is irrelevant for this part). There are 3 buses that the coaches could be on and then 10 of the other remaining 42 42 people could be on that bus. There are 10 ways to choose the other people on that bus. Then we need to pick 16 people of the remaining 32 to be on one of the other two 32 buses. There are 16 ways to do that. This leaves 16 people which must be on the last bus. We apply the multiplication principle to get: 42 32 3 × × . 10 16

(b) Same setup as in (a) but also suppose Tim Wakefield and do not want to be on the same bus. How many ways are there to distribute everyone among the three buses with 6 coaches in one bus and Wakefield and Varitek not on the same bus? Part (a) gave us total number of ways to distribute the players with 6 coaches on one bus. We count number of ways that Wakefield and Varitek are on the same bus and then (using the subtraction principle) subtract that from our answer in (a). Either Wakefield and Varitek are on the same bus as the coaches or they are not. If they are on the same bus there are 3 buses to choose from and now only 8 free spots on that bus. Then the rest of the players may go anywhere else so similar to (a) we 32 4032 have 16 options for the second bus. This gives 3 × 8 16 . If they are not on the same bus as the coach, then there are still 3 choices for the coach bus and that bus can get filled with the remaining 40 players (excluding Wakefield and 40 Varitek). This gives 3 × 10 . Then Wakefield and Varitek can be on one of 2 buses and whatever bus they are on has space for 14 more of the remaining 30 players. This 30 gives 2 × 14 . Using the addition principle, the total number of ways Wakefield and 40 30 4032 Varitek can sit on the same bus is 3 × 10 × 2 × 14 + 3 × 8 16 . 42 So altogether the total number of ways they can sit on different buses is 3 × 10 × 32 40 30 4032 16 − (3 × 10 × 2 × 14 + 3 × 8 16 ).

2 6. The season is 162 games long. Suppose the Red Sox never scored more than 13 points in a game and never had more than 11 runs scored against them. Prove that there were two games during the season with identical final scores for the Red Sox and for their opponents. (For instance, maybe the Red Sox beat two different teams 7-3.) A final baseball score consists of the number of runs the Red Sox scored and the number of runs the other team scored. If the Red Sox scored no more than 14 in any game and the opponents scored no more than 11 runs, the multiplication principle tells us that the total number of possible scores for the games are 14 × 11 = 154. Each of the 162 games (objects) must have one final score (boxes) and so if we put 162 objects into 154 boxes, the pigeonhole prinicple tells us that two objects must be in the same box. These two objects (two distinct games) will then have the same exact score.

7. At the All-Star Game this summer there were 21 starters and for the . Suppose they all had breakfast together around a circular table. How many different arrangements around the table are there? We are asking for circular permutations of 21 objects. There are P (21, 21)/21 such permutations, or 20!.

8. (a) Expand (−x + 4y)8 using the binomial theorem. n Pn n k n−k The binomial theorem says that (A + B) = k=0 k A B . Filling in for A = −x, B = 4y, and n = 6 gives 8 8 8 8 (−x)8+8(−x)7(4y)+ (−x)6(4y)2+ (−x)5(4y)3+ (−x)4(4y)4+ (−x)3(4y)5 2 3 4 5

8 8 + (−x)2(4y)6 + (−x)(4y)7 + (4y)8. 6 7

2 4 3 7 (b) What is the coefficient of x1x2x3x4 in the expansion of

16 (x1 − 2x2 − x3 + 3x4) ?

The multinomial theorem tells us that the term with exponents of variables n1 = 2, n2 = 4, n3 = 3, and n4 = 7 is   n 2 4 3 7 (x1) (−2x2) (−x3) (3x4) n1 n2 n3 n4

16! 4 3 7 and so the coefficient is 2!4!3!7! (−2) (−1) (3) .

3 n−1 n 9. Prove the formula n k−1 = k k using combinatorial reasoning. (Do not use n the formula for j directly.) Hint: Think about choosing a committee and a chair for that committee. The two sides of the equation represent two ways to do this. Suppose you have n people and you need to choose a committee with k people in it and you need a chair of that committee. One way to count the number of ways to n do this is to pick the k people for the committee ( k ) and then choose from those k n people for the chair. This is k k which is the right side. Another way to count this is to pick a chair for the committee first from the n total people (n ways to do this) and then pick the rest of the committee, so k − 1 people n−1 from the remaining n − 1. This gives n k−1 which is the left side.

4