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

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 3

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 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.

We have an infinity series appearing so for convergence we need $|z/(1+z)|\lt 1.$ The choice $\epsilon\lt 1/2$ will do.

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

This can be proven easily with a story proof similar to problem 19 in section 1.9 of introduction to probability by Blitzstein and Huang (2nd edition).

The right hand side of the expression is the number of ways of choosing a subset of size $r+s+1$ from $t+1$ elements.

Assume these $t+1$ elements are numbered and arranged in ascending order.

An alternate way of choosing the same subset is to first choose the $r+1$st element from the $t+1$ elements. Then, choose $r$ elements from the left of our chosen element. And finally, $s$ elements from the right of our chosen element.

And this alternate way of constructing the subset of size $r+s+1$ results in the left hand side.

Answer 7

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