How can I prove that $\binom{p}{0} + \binom{p+1}{1} + \binom{p+2}{2} +\dots+\binom{p+q}{q}$=$\binom{p+q+1}{q}$ using a combinatorial argument? The left part is all the permutations of $p$ white balls that have $0\le k \le q$ black balls but I don't know how I can relate this with the right part.
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2https://en.wikipedia.org/wiki/Hockey-stick_identity – Angina Seng Jul 31 '20 at 18:51
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2It suffices to realize that $\binom{p}{0}= \binom{p+1}{0}.$ – user376343 Jul 31 '20 at 18:54
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1See here https://math.stackexchange.com/questions/1490794 with interesting graphs – cgiovanardi Jul 31 '20 at 22:47
2 Answers
Ths RHS is the way of choosing $q$ objects from $p+q+1$ objects. Now,for the LHS we start from the rightmost.
We don't take the $p+q+1$th object,that gives $\binom{p+q}{q}$.
If we take it,then we are left with $p+q$ objects and $q-1$ of them to choose. Again we don't take the $p+q$th object that gives us $\binom{p+q-1}{q-1}$
And if we choose it we are left with $p+q-1$ objects and $q-2$ of them to choose.
We apply the same argument repeatedly again and again to reach the conclusion.

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Your identity is: $$\sum_{k=0}^q \binom{p+k}{k} = \binom{p+q+1}{q}$$
Rewrite as: $$\sum_{k=0}^q \binom{p+k}{p} = \binom{p+q+1}{p+1}$$
The RHS counts $(p+1)$-subsets of $\{1,\dots,p+q+1\}$. The LHS does so by conditioning on the size $p+k+1$ of the largest element. Once you have chosen that element to be the largest, choose the remaining $p$ elements from among $\{1,\dots,p+k\}$.

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