Proof of $\sum_{0 \le k \le t} {t-k \choose r}{k \choose s}={t+1 \choose r+s+1}$?

Question

How do I prove that

$$\sum_{0 \le k \le t} {t-k \choose r}{k \choose s}={t+1 \choose r+s+1} \>?$$

I saw this in a book discussing generating functions.

Answer 1

Let $r$, $t$, $s$ be fixed.

$\binom{t+1}{r+s+1}$ = number of possibilities how can I choose $r+s+1$ elements from $\{1,2,\dots,t+1\}$

Let us order the chosen elements increasingly: $a_1 < a_2 < \dots < a_{r+s+1}$

What is the number of possibilities where $a_{s+1}=k+1$? We have to choose $s$ elements from $\{1,2,\dots,k\}$ and the remaining $r$ elements from $\{k+2,\dots,t+1\}$. We have $$\binom ks \cdot \binom {t-k}r$$ possibilities.

The last expression is non-zero only for $k\ge 0$ and $t-k\ge 0$, which gives us the range of summation.

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Although you are probably not interested in a combinatorial proof, since you explicitly mentioned generating functions.

Answer 2

I'm going to make more explicit the point I think Phira is making. The identity really is just Vandermonde's convolution plus the upper negation rule for binomial coefficients.

The upper negation rule for binomial coefficients is $$\binom{n}{k} = (-1)^k \binom{k-n-1}{k},$$ which holds when $k$ is an integer (see, for example, Concrete Mathematics, 2nd edition, p. 164). Applying this, we get $$\sum_{0 \le k \le t} {t-k \choose r}{k \choose s} = \sum_{0 \le k \le t} {t-k \choose t-k-r}{k \choose k-s}= \sum_{0 \le k \le t} (-1)^{t-k-r}{-r-1 \choose t-r-k} (-1)^{k-s}{-s-1 \choose k-s}$$ $$ = (-1)^{t-r-s}\sum_{0 \le k \le t}{-r-1 \choose t-r-k}{-s-1 \choose -s+k}.$$ Now, use Vandermonde's convolution (or, more generally, the Chu-Vandermonde identity), to get $$ = (-1)^{t-r-s}{-r-s-2 \choose t-r-s},$$ and then apply the upper negation rule again, which gives us what we want: $$={t+1 \choose t-r-s} = {t+1 \choose r+s+1}.$$

Answer 3

This approach is based on generating function.

By http://math.arizona.edu/~faris/combinatoricsweb/generate.pdf

We know that $\sum \limits_{n=0}^{\infty} {n \choose k} y^n= \frac{y^k}{(1-y)^{k+1}} $

$x \cdot \sum \limits_{l=0}^{\infty} {l \choose r} x^l \cdot \sum \limits_{k=0}^{\infty} {k \choose s} x^k = x \cdot \frac{x^r}{(1-x)^{r+1}} \cdot \frac{x^s}{(1-x)^{s+1}} =\frac{x^{r+s+1}}{(1-x)^{r+s+2}} = \sum \limits_{n=0} {n \choose r+s+1} x^n$

The coefficient of $x^{t+1}$ of the $x \cdot \sum \limits_{l=0}^{\infty} {l \choose r} x^l \cdot \sum \limits_{k=0}^{\infty} {k \choose s} x^k $ is $\sum \limits_{l+k=t} {l \choose r}{k \choose s}$

Let $n=t+1$, ${t+1 \choose r+s+1} = \sum \limits_{l+k=t} {l \choose r}{k \choose s} = \sum \limits_{0 \le k \le t} {t-k \choose r}{k \choose s}$

Answer 4

Suppose we seek to verify that $$\sum_{0\le k\le t} {t-k\choose r} {k\choose s} = {t+1\choose r+s+1}.$$

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

This controls the range so we may extend $k$ to infinity, getting for the sum

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

This evaluates to $${t+1\choose t-r-s} = {t+1\choose r+s+1}$$ by inspection.

Answer 5

Note that this summation is Vandermonde's identity.

Calculate in both sums the ratio of consecutive summands and compare them. You will see that after a suitable change of variables they are the same.

Therefore, the two sums only differ by a global factor in each term and the result.

If you want to know more about this, read about hypergeometric functions.

Answer 6

Look http://fatosmatematicos.blogspot.com/2011/06/algumas-demonstracoes-da-convolucao-de.html