Simplifying this factorial expression

Question

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?

Answer 1

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:

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

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.

Answer 2

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?