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I want to prove $\sum_{k=0}^p \binom{m}{k}\binom{n}{p-k} = \binom{m+n}{p}$ for non-negative integers m,n,p. Let above statement be known as $P(m,n,p)$. I want to do induction on $p$ after fixing $m,n$.

Base case is $p=0$. LHS (left hand side) becomes $\sum_{k=0}^0 \binom{m}{k}\binom{n}{0-k} = \binom{m}{0}\binom{n}{0-0} = 1.$ And RHS (right hand side) becomes $\binom{m+n}{0} = 1.$
Hence $P(m,n,0)$ is true.

Now, I assume $P(m,n,p)$ and try to prove $P(m,n,p+1)$. So, $P(m,n,p)$ states

$\sum_{k=0}^p \binom{m}{k}\binom{n}{p-k} = \binom{m+n}{p}\binom{m}{0}\binom{n}{p} + \binom{m}{1}\binom{n}{p-1} + \cdots + \binom{m}{p}\binom{n}{p-p} = \binom{m+n}{p} \cdots\cdots (1)$.

And, I need to prove that $\sum_{k=0}^{p+1} \binom{m}{k}\binom{n}{p+1-k} = \binom{m+n}{p+1}.$

Consider the left hand side $ \binom{m}{0}\binom{n}{p+1} + \binom{m}{1}\binom{n}{p} + \cdots + \binom{m}{p+1}\binom{n}{p+1-(p+1)}.$
I tried using following binomial identity $\binom{a}{b} = \frac{a-b+1}{b} \binom{a}{b-1}$ for various terms in the sum but it lead to nowhere.

Can anybody help ?

jimjim
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user9026
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  • It's much easier to argue from a combinatorial view point. Or do you prefer to prove it by induction? In case you do, it's also easier to prove via induction on $m$ or $n$. – Sounak Jul 24 '24 at 15:52
  • @Sounak, I want to use induction. I tried using $m$ or $n$ or $p$. I am getting stuck – user9026 Jul 24 '24 at 15:58

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\begin{align} \sum_{k=0}^p \binom{m+1}{k}\binom{n}{p-k} &= \binom{m+1}{0}\binom{n}{p} +\sum_{k=1}^p \binom{m+1}{k}\binom{n}{p-k}\\ & = \binom{m+1}{0}\binom{n}{p} +\sum_{k=1}^p \binom{m}{k}\binom{n}{p-k}+\sum_{k=1}^p \binom{m}{k-1}\binom{n}{p-k}\\ & = \sum_{k=0}^p \binom{m}{k}\binom{n}{p-k}+\sum_{l=0}^{p-1} \binom{m}{l}\binom{n}{p-1-l}~\quad (l = k-1)\\ & \overset{\text{true for}~ m,n}{=} \binom{m+n}{p}+\binom{m+n}{p-1}\\ & = \binom{m+1+n}{p} \end{align}

Sounak
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