Show that $\binom{p}{0} + \binom{p+1}{1} + \binom{p+2}{2} +\dots+\binom{p+q}{q}$=$\binom{p+q+1}{q}$

Question

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.

Answer 1

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 $\{1,\dots,p+k\}$.

Answer 2

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.