Binomial Theorem Proof: (nC0)(mC0) + (nC1)(mC1) + ... + (nCm)(mCm) = (n+m C m)

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Homework Statement



To Prove:
(nC0)(mC0) + (nC1)(mC1) + ... + (nCm)(mCm) = (n+m C m)

where nC0 = n choose 0 and so on.

Homework Equations





The Attempt at a Solution


Tried expanding the whole thing using factorials - but didn't work. Any hints would be really welcome!
 
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You should review proof by induction and then try to apply it here.
 
Do I do induction on m?? So that would mean , that by assumption ...+(nCm)(mCm) = (n+m C m)...then to prove ...+(nCm+1)(m+1Cm+1) = (n+m+1 C m+1) ...correct?
 
Yes, that should work.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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