n! kn-k) (2) = 2t(rn, 2)! =n2( ) í)(n 2)(n(n 3)3):(2)2) i n(n-1)(n-2)(n-3) (2) 1 n(n-1)...

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n! kn-k) (2) = 2t(rn, 2)! =n2( ) í)(n 2)(n(n 3)3):(2)2) i n(n-1)(n-2)(n-3) (2) 1 n(n-1) n(n-1) Example 1. For n 2 2 we find = = 1. Famiarize yourself with binomials by calculating the following (some answers will be numbers, others will depend on n) 0 b) c) n-2 d) n -1

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n! kn-k) (2) = 2t(rn, 2)! =n2( ) í)(n 2)(n(n 3)3):(2)2) i n(n-1)(n-2)(n-3) (2) 1 n(n-1)...

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Add Answer to: n! kn-k) (2) = 2t(rn, 2)! =n2( ) í)(n 2)(n(n 3)3):(2)2) i n(n-1)(n-2)(n-3) (2) 1 n(n-1)...

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Add Answer to:
n! kn-k) (2) = 2t(rn, 2)! =n2( ) í)(n 2)(n(n 3)3):(2)2) i n(n-1)(n-2)(n-3) (2) 1 n(n-1)...