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Unformatted text preview: Homework 8 Solutions Section 5.3 30 a). There are C(16,5) ways to choose a committee if there are no restrictions. There are C(9,5) ways to select a committee from just the 9 men. Therefore there are C(16,5) C(9,5) = 4242 committees with at least one woman. b) There are C(16,5) ways to choose a committee if there are no restrictions. There are C(9,5) ways to select a committee from just the 9 men. There are C(7,5) ways to select a committee from just the 7 men. These two possibilities do not overlap, since there are no ways to select a committee containing neither men nor women. Therefore there are C(16,5) C(9,5) C(7,5) = 4221 committees with at least one man and at least one woman. 32 a) The only reasonable way to do this is by subtracting from the number of strings with no restrictions the number of strings that do not contain the letter a. The answer is 265 256 = 64,775,151. b)If our string is to contain both of these letters, then we need to subtract from the total number of strings the number that fail to contain one of the other (or both) of these letters. As in part (a), 256 strings fail to contain an a; similarly 256 fail to contain a b. This is overcounting, however since 246 fail to contain both of these letters. Therefore there are 256 + 256 246 strings that fail to contain at least one of the characters. Therefore the answer is 266 (256 + 256 246) = 11,737,502 c) First choose the position for the a; this can be done in 5 ways, since the b must follow it. There are four remaining positions and these can be filled in P(24,4) ways, since there are 24 letters left (no repetitions being allowed this time). Therefore the answer is 5*P(24,4) = 1,275,120 d)First choose the positions for the a and b; this can be done in C(6,2) ways, since once we pick two positions, we put the a in the leftmost and b in the other. There are four remaining positions, and these can be filled in P(24,4) ways. Therefore the answer is C(6,2) * P(24,4) = 3,825,360. 40. We can assume that the first person sits in the northernmost seat. Then there are P(5,5) ways to seat the remaining people, since they form a permutation reading clockwise from the first person. Therefore the answer is 5! = 120. Section 5.4 6) C(11,7)14 = 330. 8) C(17,9) 3829 = 81,662,929,920 10) By the binomial theorem, the typical term in this expansion is C(100,j) x(100j) (1/x)j, which can be rewritten as C(100,j)x(1002j). If we let k denote the exponent, then solving k = 1002j for j, we obtain j = (100k)/2. Thus the values of k for which xk appears in this expansion are 100,98,...2,0,2,4,...100, and for such values of k, the coefficient is C(100,(100k)/2) 28) a) To choose 2 people from a set of n men and n women, we can either choose 2 men (C(n,2) ways to do so) and 2 women (C(n,2) ways to do so) or one of each sex (n.n ways to do so). Therefore the right hand side counts the number of ways to do this (by the sum rule). The left hand side counts the same thing, since we are simply choosing 2 people from 2n people. b) 2C(n,2) + n2 = n(n1) + n2 = 2n(2n1)/2 = C(2n,2) 32. For n=0, we want (x+y)0 = C(0,j) x0j yj = C(0,0)x0y0, which is true since 1 = 1. Assume the inductive hypothesis, then we have (x+y)n+1 = (x+y) C(n,j) xnj yj = C(n,j) xn+1j yj + C(n,j) xnj yj+1 = C(n,0) xn+1+( [C(n,k) + C(n,k1)] xn+1k yk) + C(n,n)yn+1 = xn+1 + nk=1 C(n+1,k) xn+1k yk + yn+1 = n+1k=0 C(n+1,k) xn+1k yk as desired. The key point was use of Pascal's Identity to simplify the expression in brackets in the third line of this calculation. 38) We follow the hint, first noting that we can start the summation with k =1, since the term with k = 0 is 0. The lefthand side counts the number of ways to choose a subset as described in the hint by breaking it down by the number of elements in the subset; note that there are k ways to choose each of the distinguished elements if the subset has size k. For the righthand side, first note that n(n+1)2n2 = n(n1+2) 2n2 = n(n1)2n2 + n2n1 . The first term counts the number of ways to make this choice if the two distinguishable elements are different. (choose them, then choose any subset of the remaining elements to be the rest of the subset). The second term counts the number of ways to make this choice if the two distinguished elements are the same (choose it, then choose any subset of the remaining elements to be the rest of the subset). Not that this works even if n = 1. Section 5.5 4) There are 6 choices for each of 7 times, so the answer is 67 = 279,936. 32) We can treat 3 consecutive A's as one letter. Thus we have 6 letters, of which 2 are the same (the two R's), so by Theorem 3, the answer is 6!/2! = 360 42) Theorem 4 says that the answer is 52!/13!4 = 5.4 * 1028, since each player gets 13 cards. ...
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This note was uploaded on 02/12/2012 for the course EECS 203 taught by Professor Yaoyunshi during the Spring '07 term at University of Michigan.
 Spring '07
 YaoyunShi

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