piever wrote: Let $ n\ge k+1$ and $ m=n!-1$ \[ {\frac{(m+n-k)!}{m!n!}=\frac{(m+n-k)!(m+1)!}{(m+1)!m!n!}=\frac{(m+n-k)!(m+1)}{(m+1)!n!}=\frac{(m+n-k)!}{(m+1)!}\in\mathbb{Z}}\]

More general: We can choose $n$ is large enough to be greater than $k$ then let $n-k$ be $t$ with $t$ is a positive integer Then $\dfrac{(m+n-k)!}{m!n!}$ $=\dfrac{(m+t)!}{m!t![(n-k+1)(n-k+2)...n]}$ $=\binom{m+t}{t}.\dfrac{1}{(n-k+1)(n-k+2)...n}$ On the other hand $\binom{m+t}{t}=\binom{m+t-1}{t-1}.\dfrac{m+t}{t}$ So let $R=(n-k+1)(n-k+2)...n$ we can find $m$ which is big enough that $\dfrac{m+t}{t}=R.x$ then $m=Rxt-t=(Rx-1)t$ then we arrive at the infinity of values of both $m,n$ as desired and the answer is for all $k$ which is a non-negative integer (if $k=0$ then $\dfrac{(m+n-k)!}{m!n!}=\binom{m+n}{m}$ which is clearly an integer)