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Math 124 First Midterm Examination October1 1, 2010 NAME (Please Math 124 First Midterm Examination October1 1, 2010 NAME (Please print) Page Points Score 2 20 3 20 4 20 5 20 6 20 Total 100 Instructions: 1. Do all computations on the examination paper. You may use the backs of the pages if necessary. 2. Put answers inside the boxes. 3. Please signify your adherence to the honor code: I, , have neither given nor received aid in completion of this examination. 1 (20 Points) Score 1. (6 points) How many numbers between 1 and 10,000 have exactly one digit equal to 5? Take any number from 0 to 999 not containing 5 ( 93 numbers ) and insert 5 in one of 4 positions. Alternatively, the number has either 1, 2, 3 or 4 digits. If one digit, there is one choice. If two digits, either the last digit is 5 and the first is neither 0 or 5, or the last digit is not 5 and the first is 5; so there are 8 + 9 = 17 choices. If three digits, either the last digit is 5 and the middle is not 5 and the first is not 0 or 5, or the middle digit is 5 and ...; there are 8×9+8×9+9×9 = 225 choices. If four digits, there are 8 × 9 × 9 + 8 × 9 × 9 + 8 × 9 × 9 + 9 × 9 × 9 = 2673 choices. answer = 2916 2. (7 points) What is the number of ways to order the 26 letters of the alphabet so that no two of the vowels a, e, i, o and u occur consecutively? (Hint: order the consonants, then place the vowels.) There are 21! ways to order the consonants. Then there are 22 places for the first vowel. Next, there are 23 − 2 = 21 places for the next vowel to appear somewhere either than next to the first vowel. Continue in this way. 22 answer = 21! × 22!=17! = 21! × 17 × 5! 3. (7 points) Determine the number of positive integers that divide 34 ×52 ×117 ×138. Any integer divisor has the form 3i × 5j × 11k × 13` with 0 ≤ i ≤ 4 and 0 ≤ j ≤ 2 and 0 ≤ k ≤ 7 and 0 ≤ ` ≤ 8 answer = 5 × 3 × 8 × 9 = 1080 2 (20 Points) Score Pn kn n−k 1. (6 Points) Evaluate k=0(−1) k 3 . From the binomial expansion, n X n (−1)k 3n−k = (−1 + 3)n = 2n k k=0 sum = 2n 2. (7 Points) What is the probability that a permutation of the letters in the word INSTRUCTOR has three consecutive vowels? The 7 consonants are C; N; R2; S; T 2 (two are repeated). Treat the triplet of vowels as another 8 10 consonant \V". There are 2;2;1;1;1;1 ways to order the \consonants" and 2;2;1;1;1;1;1;1 to order the 10 letters in INSTRUCTOR. 8 10 probability = 3! 2;2;1;1;1;1 = 2;2;1;1;1;1;1;1 = 3!8!=10! = 1=15 3. (7 Points) How many consecutive zeros occur on the right-hand end of the decimal representation for 30! (thirty factorial)? You can get a trailing zero either by having a factor of 10, or by having a factor of 5 and a matching factor of 2. 30! involves multiplication by 10, 20 and 30, which provide 3 trailing zeros. The factors 5; 15 = 5 × 3 and 25 = 5 × 5 provide a total of four fives, with the factor 16 providing enough matching 2's. number trailing zeros = 7 3 (20 Points) Score For each of the following graphs, either redraw some edges to show that the graph is planar, or explain why the graph cannot be planar. a f e b j g h i c 1. d This is homeomorphic to K5 by col- lapsing a ! f, b ! g, c ! h, d ! i and e ! j. This is commonly known as Petersen's graph. a b g f c 2. (all vertices have degree 4) e d This is homeo- morphic to K5 by collapsing d ! b and e ! c. Alternatively, this is homeomorphic to K3;3 by deleting a $ b, b $ c, d $ e and f $ g then collapsing b to g; the result is bipartite with parts fa; d; eg and fc; f; b = gg. 4 (20 Points) Score 1. (7 Points) What is the largest possible number of vertices in a graph with 19 edges if all vertices have degree at least 3? P Since 2jEj = v2V deg(v), we have 38 = 2jEj ≥ 3jV j =) jV j ≤ 38=3 max number vertices = 12 2. (6 Points) Suppose that the graph G is formed by the 6 vertices of a cube and edges given by the 12 edges of the cube (i.e. the lines between the faces). Consider the subgraphs formed by deleting some number of the vertices of G and the edges of G that involve those vertices. What is the minimum number of vertices of G that must be deleted to produce a subgraph that is not connected? Since all vertices in the cube have degree 3, you must delete at least 3 vertices to disconnect the cube. To disconnect the cube, pick any vertex and delete its 3 neighbors. number of vertices to delete = 3 3. (7 Points) The NCAA 2010 men's basketball tournament was a single-elimination tournament (loser of any game is elminated) involving 65 teams. How many games were played to determine the champion? Each game eliminates one team; 64 teams must be eliminated. Alternatively, the competition can be represented as a binary tree with 7 levels and only two teams at the 7th level; then count the number of edges. number of games = 64 5 (20 Points) Score 1. (6 points) Is there a tree having 7 vertices, with 5 vertices of degree 2 and two vertices of degree 1? P Since jEj = jV j − 1 = 6 in a tree, and 12 = 2jEj = v2V deg(v) = 5 × 2 + 2 × 1, this is possible. In fact, the tree is a simple chain. answer = yes 2. (7 points) Is there a tree having 7 vertices, with 5 vertices of degree 1 and two vertices of degree 2? P Since jEj = 6 in a tree with 7 vertices and 12 = 2jEj = v2V deg(v) = 5 × 1 + 2 × 2 = 9 we get a contradiction. answer = no 3. (7 points) Is there a tree having 5 vertices with 2 vertices of degree 3? P Since jEj = 4 and 8 = 2jEj = v2V deg(v) ≥ 2 × 3 + 3 × 1 = 9, we get a contradiction. answer = no 6.
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