$\binom{n}{0}.\binom{m}{k}+ \binom{n}{1}.\binom{m}{k-1}+...+\binom{n}{k}.\binom{m}{0}=\binom{n+m}{k} $

Asked Mar 24 '24 at 08:50

Active Mar 24 '24 at 08:50

Viewed 28 times

I have to prove this:

$$\binom{n}{0}.\binom{m}{k}+ \binom{n}{1}.\binom{m}{k-1}+...+\binom{n}{k}.\binom{m}{0}=\binom{n+m}{k} $$ I don't know how I can prove that. I tried to use the definition or Newton Binom but I can't prove

asked Mar 24 '24 at 08:50

A12345

You have $n$ pencils and $m$ pens on a table and you want to choose $k$ of them. How many ways can you do this? – hunter Mar 24 '24 at 08:54
Note that the lower numbers always sum up to $k$. – Aig Mar 24 '24 at 08:57
https://math.stackexchange.com/q/337923/42969, https://math.stackexchange.com/q/91457/42969, https://math.stackexchange.com/q/1742596/42969, https://math.stackexchange.com/q/787440/42969 – all found with Approach0 – Martin R Mar 24 '24 at 08:58

$\binom{n}{0}.\binom{m}{k}+ \binom{n}{1}.\binom{m}{k-1}+...+\binom{n}{k}.\binom{m}{0}=\binom{n+m}{k} $

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