# sol1 - MATH 239 Assignment 1 - Suggested Solutions 1 (a) [3...

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Unformatted text preview: MATH 239 Assignment 1 - Suggested Solutions 1 (a) [3 marks] For nonnegative integers k, m, n , let A be the set of k-element subsets of { 1 , . . ., m + n } , so we have | A | = p m + n k P . But each subset α ∈ A can be written uniquely as α = β ∪ γ , where β is a subset of { 1 , . . ., m } , and γ is a subset of { m + 1 , . . ., m + n } , and | β | + | γ | = k . Therefore, if | β | = i , for some nonnegative integer i , then | γ | = k-i , and i can range from 0 to k . There are ( m i ) choices for β , of size i , and ( n k-i ) choices for γ , of size k-i , so there are ( m i )( n k-i ) choices for the pair β, γ , when β has size i . Summing over the possible values for i , we obtain that the number of choices for β, γ is k s i =0 p m i Pp n k-i P , and the identity follows, since each subset α corresponds to a unique pair β, γ . (b) [2 marks] From the binomial theorem, we know that [ x k ](1 + x ) m + n = ( m + n k ) , for each nonnegative integer k . But we also have (1 + x ) m + n = (1 + x ) m (1 + x ) n = s i ≥ p m i P x i s j ≥ p n j P x j = s i ≥ s j ≥ p m i Pp n j P x i + j , which gives [ x k ](1 + x ) m + n = k s i =0 p m i Pp n k-i P , since j = k-i , and j ≥ 0 implies that i ≤ k . We therefore have two di±erent expressions for [ x k ](1 + x ) m + n , and equating them give the identity...
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## This note was uploaded on 07/28/2009 for the course MATH math 239 taught by Professor .... during the Spring '08 term at Waterloo.

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sol1 - MATH 239 Assignment 1 - Suggested Solutions 1 (a) [3...

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