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Talk:Composition (combinatorics)

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Why not add the "more refined argument shows that the number of compositions of n into exactly k parts is given by the binomial coefficient"

Basically the same array is used but choose (k-1) places to put commas and fill the rest with plusses. This can be placed 1-1 with compositions into k parts. There are (n-1) boxes, hence the result. This is not more refined but quite understandable (A1jrj (talk) 18:31, 23 July 2010 (UTC))[reply]

This is also easily understandable from the fact that 2^k = (1+1)^k, and then just use the binomial coefficient. 77.127.178.95 (talk) 21:30, 29 August 2010 (UTC)[reply]