How do I simplify?
I was trying to simplify
$$\frac{n!k!}{k!m!(k-m)!(n-k)!}$$
and I got this, and don't know how to continues
$$\frac{n!}{m!(k-m)!(n-k)!}$$
How do I simplify n!/(m!(k-m)!(n-k)!)?
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$\begingroup$
combinatorics
discrete-mathematics
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1I don't think there is a way to make it simpler. Anyway, you can say that $\binom nk\binom km$ is an integer. – 2017-01-28
1 Answers
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We know that the number of $k$-combinations in a set of $n $ elements is given by: $$\binom {n}{k} = \frac {n!}{k!(n-k)!} $$
Thus, $$\frac {n!k!}{k!m!(n-k)!(k-m)!} = \frac {n!}{k!(n-k)!} \times \frac {k!}{m!(k-m)!} = \boxed {\binom {n}{k} \times \binom {k}{m}}$$
Hope it helps.
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0@ Rohan But I was asked to to the reverse. that's why I translate the binomial notation as you showed, into the fractions. I need to show $$\sum_{k=m}^{n}\binom{k}{m}\binom{n}{k}$$ in a simple formula not containing a sum. – 2017-01-28
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0@Tmm That is a different question from what you asked. If you had provided that context then we would have a better idea what you were looking for. But you only said "simplify", and this is unarguably simplified. There are plenty of identities that allow simplification of sums of binomial coefficients. But you didn't mention anything about a sum. – 2017-01-28