55PS6 - SOLUTION SET 6—DUE 3/12/2008 This solution set is...

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Unformatted text preview: SOLUTION SET 6—DUE 3/12/2008 This solution set is currently only partially complete, to allow a post before the midterm. Please report any errors in this document to Ian Sammis ( isammis@math.berkeley.edu ). Problem 1 (#5.3.12) . How many bit strings of length 12 contain a) exactly 3 1s? b) at most three 1s? c) at least thee 1s? d) an equal number of 0s and 1s? Solution. In each case, once you know that you want m 1s, there are ( 12 m ) ways to arrange those 1s (you choose the positions of the 1s). Thus a) If there are exactly 3 1s, then there are ( 12 3 ) = 220 ways to place them into the bit string. Since this is a bit string, once the 1s have been placed the string is fully specified. b) Now, there are either 0, 1, 2, or 3 ones. Since those cases are disjoint, we can sum the counts, for ( 12 ) + ( 12 1 ) + ( 12 2 ) + ( 12 3 ) = 1+12+66+220 = 299 such strings. c) If there are at least three 1s, we’d have to sum over a lot of cases to use the technique of part (b). It’d be much easier to subtract off the cases that we aren’t interested in. There are 2 12 = 4096 bit strings of length 12, of which we don’t care about the ones with 0 ones (1 case), 1 one (12 cases), or 2 ones (66 cases). Thus there are 4096- 1- 12- 66 = 4017 bit strings of length 12 with at least three 1s. d) If there are an equal number of 0s and 1s, there are 6 of each. There are ( 12 6 ) = 924 such bit strings. Problem 2 (#5.3.16) . How many subsets with an odd number of elements does a set with 10 elements have? Solution. We have to choose 1,3,5,7, or 9 elements of the 10 to be in our subset. There are ( 10 1 ) + ( 10 3 ) + ( 10 5 ) + ( 10 7 ) + ( 10 9 ) ways to do this. We can save ourselves a little bit of work by using the symmetry of the binomial coefficients to write this as 2 ( 10 1 ) + 2 ( 10 3 ) + ( 10 5 ) = 2(10) + 2(120) + 252 = 512 . Problem 3 (#5.3.26) . Thirteen people on a softball team show up for a game. a) How many ways are there to choose 10 players to take the field?...
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This note was uploaded on 12/07/2011 for the course MATH MATH192 taught by Professor Michealalain during the Spring '11 term at Université Joseph Fourier Grenoble I.

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55PS6 - SOLUTION SET 6—DUE 3/12/2008 This solution set is...

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