Ytukyg

I need to prove that $frac{(m+n-1)!}{m!n!}$ is an integer there is a question here which is similar to this question but i cant figure out how I can insert the '-1' in there. Thanks

Since
$$frac{(m+n-1)!}{m!n!}=frac{(m+n)}{(m+n)}frac{(m+n-1)!}{m!n!}=frac{1}{m+n}binom{m+n}{m}.$$
So we need to show that $frac{1}{m+n}binom{m+n}{m}$ is an integer. Observe that $gcd(m,n)=1$ also implies that $color{blue}{gcd(m+n,m)=1}$. Since $gcd(a,b)$ is a linear combination of $a$ and $b$. Therefore,
begin{align*}
frac{1}{m+n}binom{m+n}{m} &=frac{color{blue}{gcd(m+n,m)}}{m+n}binom{m+n}{m}\
&=frac{color{blue}{x(m+n)+y(m)}}{m+n}binom{m+n}{m} &text{for some } x,y in Bbb{Z}\
&=xleft[color{green}{frac{m+n}{m+n}binom{m+n}{m}}right]+yleft[color{red}{frac{m}{m+n}binom{m+n}{m}}right].
end{align*}

The quantity $color{green}{frac{m+n}{m+n}binom{m+n}{m}}=binom{m+n}{m}$ is definitely an integer and the quantity
$$color{red}{frac{m}{m+n}binom{m+n}{m}}=binom{m+n-1}{m-1}$$
is also an integer.

This implies that $frac{1}{m+n}binom{m+n}{m}$ is an integer combination of these two integers, hence an integer itself.

Hint: This is equivalent to showing that $m+n$ divides $binom{m+n}{m}$, or equivalently $a$ divides $binom{a}{b}$ for any $a>b$ where $gcd(a,b)=1$. See if you can partition the subsets of $a$ objects with $b$ elements into blocks of $a$ to prove this combinatorially. If you know it, you might want to recall the "necklace proof" of Fermat's Little Theorem - this has a similar flavor.