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The following input

   (n ∈ Integers) && (n > 0) && (k ∈ Integers) && (k >= 0) && (k <= n), 
   FullSimplify[Binomial[n, k]/Binomial[n, k + 1]]



Why does it not give


and how do I make it do this?

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up vote 11 down vote accepted

Try using FunctionExpand:

FunctionExpand[Binomial[n, k]/Binomial[n, 1 + k]]    
  (1 + k)/(-k + n)

Mathematica's understanding of what is simple is based on leaf count and can be unintuitive at times.

share|improve this answer
perfect, thanks Searke – Lucas Nov 29 '12 at 0:24
LeafCount indeed gets you a complexity of 11 for both terms. Using the written size of the expression with ComplexityFunction -> (StringLength@ToString[#] &)] still yields the same result, though FullSimplify now delivers the version that the OP wants. – Sjoerd C. de Vries Nov 29 '12 at 16:05
Somehow, I have never considered StringLength as a ComplexityFunction until now. – Searke Nov 29 '12 at 16:06

Another way to proceed would be to make Binomial more expensive (or less simple) using a ComplexityFunction:

f[e_] := 100 Count[e, _Binomial, {0, Infinity}] + LeafCount[e]

FullSimplify[Binomial[n, k]/Binomial[n, k + 1], ComplexityFunction -> f]


(1 + k)/(-k + n)
share|improve this answer
Sounds, erm, complex. But it is interesting and good to know. – Lucas Nov 29 '12 at 0:32
It is interesting, that Simplify[Binomial[n, k]/Binomial[n, 1 + k], ComplexityFunction -> f] with your definition for f gives Binomial[n, k]/Binomial[n, 1 + k]. It looks like Simplify does not try enough transformations of Binomial. – au700 Nov 29 '12 at 0:34
Probably. You could perhaps find out using another recent Q+A of mine… – WalkingRandomly Nov 29 '12 at 0:39
@WalkingRandomly It is overkill 100 Count..., 1 Count... is good enough, +1. – Artes Nov 30 '12 at 0:43

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