By a combinatorial argument, prove that for $\displaystyle r \leq n$ and $\displaystyle r \leq m$, $\displaystyle \binom {n+m} {r} = \binom {m} {0} \binom {n} {r} + \binom {m} {1} \binom {n} {r-1} + ... + \binom {m} {r} \binom {n} {0}$.

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- Sep 30th 2010, 12:29 AMZennieCombinatorial proof
By a combinatorial argument, prove that for $\displaystyle r \leq n$ and $\displaystyle r \leq m$, $\displaystyle \binom {n+m} {r} = \binom {m} {0} \binom {n} {r} + \binom {m} {1} \binom {n} {r-1} + ... + \binom {m} {r} \binom {n} {0}$.

- Sep 30th 2010, 12:33 AMaman_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)?