show that ${r \choose 0}{s \choose n}+{r \choose 1}{s \choose n+1}+{r \choose 2}{s \choose n+2}+...+{r \choose n}{s \choose n+n}={r+s \choose s-n}$

Question

show that $${r \choose 0}{s \choose n}+{r \choose 1}{s \choose n+1}+{r \choose 2}{s \choose n+2}+...+{r \choose n}{s \choose n+n}={r+s \choose s-n}$$

I tried the counting argument, but the total items selected in each term is different, so that doesn't seem to work. The equation resembles the formula

$${r \choose 0}{s \choose n}+{r \choose 1}{s \choose n-1}+...+{r \choose n}{s \choose 0}={r+s \choose n}$$

but I have no idea how to take advantage of it

Answer 1

Your question is wrong. Note that if we set $r=2, s=3, n=1$, then $$\binom{2}{0}\binom{3}{1}+\binom{2}{1}\binom{3}{1+1}=9 \neq 10=\binom{3+2}{3-1}$$ So, we cannot prove it as it is incorrect.