Combinatorial Identity with Binomial Coefficients: $ {{a+b+c-1}\choose c} = \sum_{i+j=c} {{a+i-1}\choose i}{{b+j-1}\choose j} $

Question

I got the following identity from commutative algebra.

I am curious to see elegant elementary methods.

$$ {{a+b+c-1}\choose c} = \sum_{i+j=c} {{a+i-1}\choose i}{{b+j-1}\choose j} $$

Answer 1

It is just a case of Vandermonde's identity.

The LHS is the coefficient of $x^c$ in $\frac{1}{(1-x)^{a+b}}$, by stars and bars.

On the other hand, $\frac{1}{(1-x)^{a+b}}=\frac{1}{(1-x)^{a}}\cdot\frac{1}{(1-x)^b}$, hence:

$$\begin{eqnarray*} \binom{a+b+c-1}{c}=[x^c]\frac{1}{(1-x)^{a+b}}&=&\sum_{d=0}^{c}[x^d]\frac{1}{(1-x)^a}\cdot [x^{c-d}]\frac{1}{(1-x)^b}\\ &=& \sum_{i+j=c}\binom{a+i-1}{i}\binom{b+j-1}{j}, \end{eqnarray*}$$ as wanted.

Answer 2

For variety's sake here is a slightly different approach. Suppose we seek to evaluate

$$\sum_{k=0}^c {k+a-1\choose a-1} {b-1+c-k\choose b-1}.$$

Introduce $${b-1+c-k\choose b-1} = {b-1+c-k\choose c-k} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{c-k+1}} (1+z)^{b-1+c-k} \; dz.$$

Observe that this is zero when $k\gt c$ so we may extend $k$ to infinity to get for the sum

$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{c+1}} (1+z)^{b-1+c} \sum_{k\ge 0} {k+a-1\choose a-1} \frac{z^k}{(1+z)^k} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{c+1}} (1+z)^{b-1+c} \frac{1}{(1-z/(1+z))^a} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{c+1}} (1+z)^{a+b-1+c} \frac{1}{(1+z-z)^a} \; dz \\ = {a+b+c-1\choose c}.$$

We can take $\epsilon \lt 1/2$ for the series to converge.

Answer 3

Using the identity $$ \sum_{j=0}^n\binom{n-j}{k}\binom{j}{m}=\binom{n+1}{k+m+1} $$ proven in this answer, we get $$ \begin{align} \sum_{i+j=c}\binom{a+i-1}{i}\binom{b+j-1}{j} &=\sum_{i+j=c}\binom{a+i-1}{a-1}\binom{b+j-1}{b-1}\\ &=\binom{a+b+c-1}{a+b-1}\\ &=\binom{a+b+c-1}{c}\\ \end{align} $$

Answer 4

Rewrite as $$\binom{a+b+c-1}{a+b-1} = \sum_{i=0}^c \binom{a+i-1}{a-1}\binom{b+c-i-1}{b-1},$$ which we can prove combinatorially by counting the number of $(a+b-1)$-subsets of $\{1,\dots,a+b+c-1\}$ in two different ways. The LHS is clear. To obtain the RHS, condition on the $a$th smallest element $a+i$. Then we must choose $a-1$ smaller elements from $\{1,\dots,a+i-1\}$ and the remaining $b-1$ larger elements from $\{a+i+1,\dots,a+b+c-1\}$.

Answer 5

Essentially, Marko Riedel's comment above. $$ \begin{split} \sum_{\substack{i+j=c\\i\ge 0,j\ge 0}}\binom{a+i-1}{i}\binom{b+j-1}{j} &=\sum_{\substack{i+j=c\\i\ge 0,j\ge 0}}(-1)^i\binom{-a}{i}(-1)^j\binom{-b}{j}\\ &=\sum_{\substack{i+j=c\\i\ge 0,j\ge 0}}(-1)^{i+j}\binom{-a}{i}\binom{-b}{j}\\ &=(-1)^c\binom{-a-b}{c}=\binom{a+b+c-1}{c} \end{split} $$