# Combinatorial proof

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• September 30th 2010, 01:29 AM
Zennie
Combinatorial proof
By a combinatorial argument, prove that for $r \leq n$ and $r \leq m$, $\binom {n+m} {r} = \binom {m} {0} \binom {n} {r} + \binom {m} {1} \binom {n} {r-1} + ... + \binom {m} {r} \binom {n} {0}$.
• September 30th 2010, 01:33 AM
aman_cc
Hint - To select r objects from (n+m) objects - break (n+m) objects in two groups (n) and (m) objects. How will you go about selecting r objects in total now (from both the groups)?