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Unformatted text preview: MATH 55 Prof. Bernoff Fall 2007 Harvey Mudd College Homework 3 Solutions 17.6 Prove: n k = k + 1 n 1 . Solution: There is a bijective argument that these two are equal using the idea of two arrangements that are duals that is two arrangements that can be made into each other by exchanging how objects are labelled. The number of ways to arrange k bars and n 1 stars is the same way as the number of ways to arrange k stars and n 1 bars. By the stars and bars argument, (( n k )) counts the number of ways to place k stars around n 1 bars. Similarly, (( k +1 n 1 )) counts the number of ways to place n 1 stars around k bars, and so theyre equal. 17.7 Let hh n k ii denote the number of multisets of cardinality k we can form choosing the elements in { 1 , 2 , 3 ,...,n } with the added condition that we must use each of these n elements at least once in the multiset. (a) Evaluate from first principles, n n . Solution: There is only one way to pick a set of n elements from { 1 , 2 , 3 ,...,n } with each number represented at least once. (b) Prove: n k = ( n k n ) . Solution: After choosing each of 1 ,...,n for a multiset of size k , we choose the remaining k n elements as a multiset from { 1 , 2 , 3 ,...,n } , so n k = (( n k n )) . 17.8 Let n,k be positive integers. Prove: n k = n 1 + n 1 1 + n 1 2 + n 1 k Solution: There are (( n k )) kmultisets of { 1 , 2 , 3 ,...,n } , but we can count them all by conditioning on how many times n appears in each multiset. The kmultisets in which n appears i times are the (( n 1 k i )) ( k i )multisets of { 1 , 2 , 3 ,...,n 1 } (which fill in the rest of each kmultiset). Since i can range from 0 to k , we have (( n k )) = k i =0 (( n 1 k i )) . 17.9 Let n,k be positive integers. Prove: n k = 1 k 1 + 2 k 1 + 3 k 1 + n k 1 Solution: We count the kmultisets of { 1 , 2 , 3 ,...,n } by conditioning on the largest element of each. If i is the largest element in a kmultiset of { 1 , 2 , 3...
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 Fall '07
 Bernoff
 Math

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