Talk:Number of Compositions

Can $n$ legitimately be zero in this case? If it is, we need to treat it as a separate case as $\binom {n-1}{k-1}$ is wrong when $n = 0$. IMO better to state that $n > 0$ (i.e. is strictly positive). --prime mover 06:41, 6 January 2011 (UTC)

Well since there aren't $2^{0-1}=\frac12$ total compositions of 0, I would say it doesn't hold for $n=0$... --Alec (talk) 21:02, 9 January 2011 (UTC)