While looking at the solution of a question, I realised that it used something like this in one of its steps without any prior mention of the fact which makes me think either this is something too obvious that I'm missing or it is a wide known generalized theorem of some sort that I don't know of. Either way, please help me out.
$\sum_{r=0}^n{\binom{n}{n-r}\binom{2n}r}= \binom{3n}n$
How is that?
Suppose in box ,there are total $3n$ balls ,of which $n$ are red and $2n$ are blue. And we want to select $n$ ball out of $3n$ balls, this can be done in $\binom{3n}{n}$ ways.
But there is another way of counting:
We can select $n$ red ball and $0$ blue balls in $\binom{n}{n} \binom{2n}{0}$ $\text{or}$ $n-1$ red balls and $1$ blue balls in $\binom{n}{n-1} \binom{2n}{1}$ $\text{or}.......$ $(n-n)$ red ball and $n$ blue balls in $\binom{n}{n-n} \binom{2n}{n}$
$=\binom{n}{n} \binom{2n}{0}+\binom{n}{n-1} \binom{2n}{1}+.....+\binom{n}{n-n} \binom{2n}{n}$
$=\sum_{r=0}^n{\binom{n}{n-r}\binom{2n}r}$