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I know that $\frac{(m-1)!}{(m-n)!(n-1)!} + \frac{(m-1)!}{(m-n-1)!(n)!} = \frac{m!}{(n)!(m-n)!}$, but I am not sure on the intermediate steps. The only solution I am seeing involves finding a common denominator:

$\frac{(m-1)!}{(m-n)!(n-1)!} + \frac{(m-1)!}{(m-n-1)!(n)!} = \frac{(m-1)!(m-n-1)!(n)!+(m-1)!(m-n)!(n-1)!}{(m-n)!(n-1)!(m-n-1)!(n)!}$

But I don't really see where to go from there. Is there a more simple way to go about this? If not, how do I simplify the monster expression above?

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Try to make the two denominators on the left look like the one on the right. So for the first expression on left, multiply top and bottom by $n$. For the second one, multiply top and bottom by $m-n$. – André Nicolas Oct 5 '11 at 22:10
up vote 3 down vote accepted

For first part: $\frac{(m-1)!}{(m-n)!(n-1)!}$

we want to write (n-1)! as n!, since n!=n*(n-1)!, we need to multiply by [n] to get:


which is: $\frac{n*(m-1)!}{(m-n)!(n)!}$ --> (a)

For second part: $\frac{(m-1)!}{(m-n-1)!(n)!} $ we want to write (m-n-1)! as (m-n)!, since (m-n)!=(m-n)*(m-n-1)! we need to multiply by [m-n] to get:

$\frac{[m-n](m-1)!}{[m-n](m-n-1)!(n)!} $

which is: $\frac{(m-n)(m-1)!}{(m-n)!(n)!} $

In the above expression, m(m-1)! is just m! so we can write that part as:

$\frac{m!-n(m-1)!}{(m-n)!(n)!} $ --> (b)

Now, you can add expression (a) to expression (b) to get the desired result.

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Your common denominator is very complex. Why don't you pair off some of the many common factors in the factorials and use $(m-n)!n!$ as a common denominator instead?

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How do I simplify the factorials like that? – KPthunder Oct 5 '11 at 22:03
For example, $(m-n)(m-n-1)!n!=(m-n)!n!=$ by definition of $(m-n)!$. – Henning Makholm Oct 5 '11 at 22:05

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