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I encountered the following question in a discrete math course:

Prove that $ \binom{2n}{k-1} < \binom{2n}{k} $ for $k = 1, 2, \ldots , n$.

Hint: This should be a very cleanly written proof.

I'm working through this proof and I'm at a step where I have similar numerators but with a factorial.

My question is: Can I "cancel out" factorials?

Yuval Filmus
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Cintra
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  • Well, factorials are simply product of integers, and if you have the same set of integers in numerator and denominator, you can cancel them, so... – rogerl Mar 27 '15 at 00:26
  • Do you need a proof for that combinatorial identity? If so, then I can write one. – Prasun Biswas Mar 27 '15 at 00:30

2 Answers2

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You can cancel, but you have to be careful how you do it. For example: $\frac{k!}{(k+1)!}$ can be reduced to $\frac{1}{k+1}$ by cancelling the factorial portion on top, but you CANNOT cancel like: $\frac{k!}{(k+1)!}=\frac{k}{k+1}$ effectively cancelling the "!" symbol.

Think about what the factorial means and it will be clear what you can do.

$$\frac{k!}{(k+1)!}=\frac{k(k-1)(k-2)...(3)(2)(1)}{(k+1)(k)(k-1)(k-2)...(3)(2)(1)}=\frac{1}{k+1}.$$

Other factorials are similar. Handy things to keep in mind are facts like $(n+1)!=(n+1)\cdot n!$, etc.

TravisJ
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Hint: After you cancel out factorials, you will find out that $$ \frac{\binom{2n}{k}}{\binom{2n}{k-1}} = \frac{2n-k+1}{k} > 1. $$

Yuval Filmus
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