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]

Lacks a definition

There is no definition of what "a k-composition of n" is in the last section before the references. Propose this to be "A k-composition of n is a composition of n with at most k elements".Bj norge (talk) 18:01, 14 April 2011 (UTC)[reply]

Inaccurate statement

The statement "This is a power of two, because every composition matches a binary number" is not quite correct. The "matches" have to exhaust (the binary representations of) the integers in the set {0, 1, ..., 2^(n-1) - 1} or some other set with cardinality 2^(n-1). Ed Jeffery (talk) 06:41, 16 May 2013 (UTC)[reply]