Are these combinatorial expressions equivalent?

Question

I am trying to figure out whether the following expressions are equivalent: $$\sum_{i=0}^{n}\binom{i}{a+b} = \sum_{i=0}^{n}\binom{i}{a}\binom{n-i}{b}$$ It seems intuitively true to me and it's held up for several test values but I can't figure out how to prove it combinatorially. Are the expressions equivalent? And if so why?

Answer 1

They are both equal to $\binom{n+1}{a+b+1}$: both count the $(a+b+1)$-element subsets of $S=\{0,1,\ldots,n\}$.

Let $A$ be such a subset; there is a unique $i_A\in A$ such that $a$ members of $A$ are less than $i_A$, and $b$ members of $A$ are greater than $i_A$. For a given value of $i\in S$, there are $\binom{i}a$ ways to choose $a$ smaller members of $S$ and $\binom{n-i}b$ ways to choose $b$ larger members of $S$, so there are $\binom{i}a\binom{n-i}b$ subsets $A$ such that $i_A=i$. Summing over $i$ yields the identity

$$\sum_{i=0}^n\binom{i}a\binom{n-i}b=\binom{n+1}{a+b+1}\;.$$

Alternatively, we can count the $(a+b+1)$-element subsets of $S$ by their largest elements: for each $i\in S$ there are $\binom{i}{a+b}$ such subsets with largest element $i$. Thus,

$$\sum_{i=0}^n\binom{i}{a+b}=\binom{n+1}{a+b+1}$$

as well, and your identity is established.