I tried to do this exercise from a guide of my gf, but I couldn't. The exercise is: $$\sum_{i=0}^{n}\binom{n}{i}^{2} = \frac{(2n)!}{n!n!} $$
If anyone can help me, I would really appreciate it!
I used Pascal's rule and a lot of decompositions, but I couldn't, so I think that isn't really necessary put my work. I need the correct way or a good idea :D thanks!
On
Imagine there are two group of people, with each group $n$ people. Now, we want to choose $n$ people from these $2n$ people. On one hand, the number of combinations we can have is $\sum^n_{i=0} \binom{n}{i} \binom{n}{n-i}=\sum^n_{i=0} \binom{n}{i}^2$, since we can choose $i$ from one group and $n-i$ from another group. On the other hand, we can neglect the groups, and it is just $\binom{2n}{n}=\frac{2n!}{n!n!}$.
This is the well known Vandermonde's identity. There's a rather nice proof of a generalization of it. Consider the generating functions $(1 + x)^n\cdot(1 + x)^m$ and $(1 + x)^{n + m}$. Then obviously the $r$th degree term must be the same for both generating functions. Expand the first one to get $$ \left(1 + \binom n 1 x + \cdots + \binom n n x^n\right)\cdot\left(1 + \binom{m}{1} x + \cdots + \binom {m}{m}x^m\right).$$ Then the term of degree $r$ has coefficient $$ \binom n r + \binom n {r-1}\binom{m}1 + \binom n {r-2}\binom{m}2 +\cdots+\binom{m}{r}$$ This is equal to the term of degree $r$ of the second generating function, which is $\binom{m+n}{r}$.