Can't figure out a step in a binomial theorem example

Question

today I was practicing math and I've come to a example where I can't figure out how did they got this result. Here you have the example:

and this is the step which I don't get:

Answer 1

Let $S=\{0,1,\ldots,2n\}$. Let $\mathscr{A}_k$ be the family of subsets $A$ of $S$ such that

• $|A|=n+1$;
• $k+m\in A$; and
• $|\{a\in A:a<k+m\}|=k$.

Because it contains $0$, $S$ has $k+m$ elements less than $k+m$, so there are $\binom{k+m}k$ ways to choose the $k$ elements of $A$ smaller than $k+m$. There are $2n-(k+m)$ elements of $S$ larger than $k+m$, so there are $\binom{2n-(k+m)}{n-k}$ ways to choose the rest of $A$. Thus,

$$|\mathscr{A}_k|=\binom{k+m}k\binom{2n-(k+m)}{n-k}=\binom{k+m}k\binom{2n-k-m}{n-m}\;,$$

since $2n-k-m-(n-k)=n-m$. $|S|=2n+1$, so $S$ has $\binom{2n+1}{n+1}$ subsets of cardinality $n+1$, and we have

$$\sum_k\binom{k+m}k\binom{2n-k-m}{n-m}=\sum_k|\mathscr{A}_k|=\binom{2n+1}{n+1}\;.$$

Finally,

$$\begin{align*} \frac{(n!)^2}{(2n)!}\sum_k\binom{k+m}k\binom{2n-k-m}{n-m}&=\frac{(n!)^2}{(2n)!}\binom{2n+1}{n+1}\\ &=\frac{(n!)^2(2n+1)!}{(2n)!(n+1)!n!}\\ &=\frac{2n+1}{n+1}\;. \end{align*}$$