I am trying to solve the following problem.
If $\displaystyle{22\choose0}{15\choose10}+{22\choose1}{15\choose9}+{22\choose2}{15\choose8}+\cdots+{22\choose9}{15\choose1}+{22\choose10}{15\choose0}={a\choose b}$, compute all possible values of $a+b$ where $a>b>0, b \neq 1$.
I tried writing it using $${n \choose k}=\frac{n!}{k!(n-k)!}$$ and seeing if anything would cancel, but the work got messy. I can't think of any identities that can be applied here either.
What identities or manipulations can be applied to solve this? Thanks!
This is just an example of Vandermonde's identity.
The given expression can be written as $$\sum_{i=0}^{10}\binom {22}{i} \binom {15}{10-i}$$ Hence by using Vandermonde's identity we get $$\sum_{i=0}^{10}\binom {22}{i} \binom {15}{10-i}$$ $$=\binom {37}{10}$$ $$= \binom {37}{27}$$ $$= \binom {348330136}{348330135} $$
You can visit
https://artofproblemsolving.com/wiki/index.php?title=Combinatorial_identity
For more information on Vandermonde's identity.