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Unformatted text preview: Solutions to Homework 6 Section 5.1 12. We use the sum rule, adding the number of bit strings of each length up to 6. If we include the empty string, then we get 2 + 2 1 + 2 2 + 2 3 + 2 4 + 2 5 + 2 6 = 2 7 1 = 127 . 24. (a) There are 10 ways to choose the first digit, 9 ways to choose the second, and so on; therefore the answer is 10 · 9 · 8 · 7 = 5040 . (b) There are 10 ways to choose each of the first three digits and 5 ways to choose the last; therefore the answer is 10 3 · 5 = 5000 . (c) There are 4 ways to choose the position that is to be different from 9, and 9 ways to choose the digit to go there. Therefore there are 4 · 9 = 36 such strings. 34. There are 2 n such functions, since there is a choice of 2 function values for each element of the domain. 52. We draw the tree, with its root at the top. We show a branch for each of the possibilities 0 and 1, for each bit in order, except that we do not allow three consecutive 0’s. Since there are 13 leaves, the answer is 13. Section 5.2 4. We assume that the woman does not replace the balls after drawing them. (a) There are two colors: these are the pigeonholes. We want to know the least number of pigeons needed to insure that at least one of the pigeonholes contains three pigeons. By the generalized pigeonhole principle, the answer is 5. 1 Figure 1: Section5.1 Exercise 52. (b) She needs to select 13 balls in order to insure at least three blue ones. If she does so, then at most 10 of them are red, so at least three are blue. 18. (a) If not, then there would be 4 or fewer male students and 4 or fewer female students, so there would be 4 + 4 = 8 or fewer students in all, contradicting the assumption that there are 9 students in the class. (b) If not, then there would be 2 or fewer male students and 6 or fewer female students, so there would be 2 + 6 = 8 or fewer students in all, contradicting the assumption that there are 9 students in the class. 22. This follows immediately from Theorem 3, with n = 10 . 40. Look at the pigeonholes { 1000 , 1001 } , { 1002 , 1003 } , . . . , { 1098 , 1099 } . There are clearly 50 sets in this list. By the pigeonhole principle, if we have 51 numbers in the range from 1000 to 1099 inclusive, then at least two of them must come from the same set. These are the desired two consecutive house numbers....
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This note was uploaded on 06/07/2011 for the course COT 3100 taught by Professor Staff during the Summer '08 term at University of Florida.
 Summer '08
 Staff

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