ma017-2010-sol5

ma017-2010-sol5 - MA017 2010 HOMEWORK 5 (1) One can prove (...

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Unformatted text preview: MA017 2010 HOMEWORK 5 (1) One can prove ( n k ) + ( n k +1 ) = ( n +1 k +1 ) by “pure thought” as follows. The ( k + 1)-element subsets of { 1 ,...,n + 1 } are of two mutually exclusive types: those that contain n + 1, and those that do not; the former are in bijection with the k-element subsets of { 1 ,...,n } , the latter with the ( k + 1)-element subsets of { 1 ,...,n } . Give similar proofs for whichever of the following combinatorial identities strike your fancy. Assume, in all formulae where ( n k ) appears, that k ≤ n . (a) ( n k ) = ( n n- k ) . Choosing a k-subset of an n-set is equivalent to choosing its complement, which has n- k elements. (b) ∑ n k =0 ( n k ) = 2 n . Both sides count the number of subsets of an n-set. One can partition the subsets of an n-set according to their size k . Summing over the number of k-subsets for each size k , one gets the number of subsets of an n-set. Alternatively, a subset of an n-set is determined by the membership of each of its n elements, so there are 2 n possibilities. (c) ∑ n k =0 k ( n k ) = n 2 n- 1 . Both sides count the number of ways to choose a subset of an n-set with one marked element. For each subset we choose, of size say k , there are k elements we can mark. Alternatively, we can choose the marked element first, and then there are 2 n- 1 choices for the unmarked part of the subset. (d) ∑ n k =0 k 2 ( n k ) = n 2 n- 1 + n ( n- 1)2 n- 2 . Both sides count the number of ways to choose a subset of an n-set with two distinct markings, possibly of the same element. We can choose our subset first, and then choose which of the two elements to mark, giving the LHS. Alternatively, we can choose the markings first. In the n cases in which they are applied to the same element there are 2 n- 1 possibilities for the unmarked part of the subset, while in the n ( n- 1) cases in which they are applied to distinct elements there are 2 n- 2 choices for the remaining unmarked part of the subset. (e) ∑ n k =0 k 3 ( n k ) = n 2 n- 1 + 3 n ( n- 1)2 n- 2 + n ( n- 1)( n- 2)2 n- 3 . Both sides count the number of ways to choose a subset of an n-set with three distinct markings, possibly of the same elements. We can choose our subset first, and then choose which of the three elements to mark, giving the LHS. Alternatively, we can choose the markings first. In the n cases in which they are applied to the same element there are 2 n- 1 possibilities for the unmarked part of the subset, in the 3 n ( n- 1) cases in which they are applied to two distinct (there are n ( n- 1) 2 choices for the pair of elements to mark, and 6 ways to mark them) there are 2 n- 2 choices for the remaining unmarked part of the subset, while in the n ( n- 1)( n- 2) cases in which all three marked elements are distinct, 2 n- 3 possibilities remain....
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ma017-2010-sol5 - MA017 2010 HOMEWORK 5 (1) One can prove (...

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