Thread: Binomial Theorem Problem
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Old August 27th, 2008, 07:01 AM
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Quote:
Originally Posted by Number Cruncher 20 View Post
By any chance would you know how to get started on proving that:

for .
This is straightforward. We have

\binom{n}{k}\binom{k}{l}=\frac{n!}{k!(n-k)!}\cdot \frac{k!}{l!(k-l)!} and \binom{n}{l}\binom{n-l}{k-l}=\frac{n!}{l!(n-l)!}\cdot \frac{(n-l)!}{(k-l)!(n-l-(k-l))!}

If you simplify these two expressions you'll see that they are equal.
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