Proving the identity $ \sum_{k=0}^{\min\{p,q\}} \binom{n}{k} \binom{n-k}{p-k} \binom{n-p}{q-k} = \binom{n}{p} \binom{n}{q}$

Question

While preparing for an exam, I am trying to prove the identity below. I searched for similar questions but only found Combinatorial proof of ${n\choose p}{n\choose q}=\sum_{k=0}^n {n \choose k}{n-k \choose p-k}{n-k \choose q-k}$ (Someone mentioned that this identity may be not true).

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Question

Let $p,q ,n\in \mathbb{N}$ s.t $p,q \le n$

$$ \sum_{k=0}^{\min\{p,q\}} \binom{n}{k} \binom{n-k}{p-k} \binom{n-p}{q-k} = \binom{n}{p} \binom{n}{q}$$

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My attempt

• First of all, I want to find a combinatorial proof since it's important for me to understand the "combinatorial story" behind it.

• I draw this in order to understand maybe what's the "story". But I'm stuck understanding what are the purple elements and the whole connection to the LHS (Note that the index k "runs" over vaules from $0$ to $\min{(p,q)}$).

I would appreciate if someone here can try to continue my solution or suggest an alternative solution (if you do please explain each step the motivation behind it).

Answer 1

Disclaimer

Just a summary of your own attempt, comment by J.G., and answer by Robert Wegner

Set Up

There are $n$ people, choose $p$ to play football and choose $q$ to play volleyball. Some people may play both of them.

Right Hand Side

• Choose $p$ out of $n$ to play football
• Choose $q$ out of $n$ to play volleyball

The total number of possibilities is given by the right hand side.

Left Hand Side

• Choose $k\in\left[0,\min{\left(p,q\right)}\right]$ to play both
• From remaining $n-k$, choose $p-k$ to play only football
• From remaining $n-p$, choose $q-k$ to play only volleyball

Summing up for all allowed value of $k$, the total number of possibilities is then given by the LHS.

Conclusion

Since both count the same object, RHS must equal LHS.

Answer 2

Since OP is also asking for alternate solutions here is a variation based upon hypergeometric functions. We use the rising factorial notation $(a)_{k}:=a(a+1)\cdots(a+k-1)$.

We obtain \begin{align*} \color{blue}{\sum_{k=0}^{\min\{p,q\}}}&\color{blue}{\underbrace{\binom{n}{k}\binom{n-k}{p-k}\binom{n-p}{q-k}}_{=:t_k}}\\ &=\sum_{k=0}^{\min\{p,q\}}t_k=t_0\sum_{k=0}^{\min\{p,q\}}\prod_{j=0}^{k-1}\frac{t_{j+1}}{t_j}\\ &=\binom{n}{p}\binom{n-p}{q}\sum_{k=0}^{\min\{p,q\}}\prod_{j=0}^{k-1}\binom{n}{j+1}\binom{n-j-1}{p-j-1}\binom{n-p}{q-j-1}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\cdot\binom{n}{j}^{-1}\binom{n-j}{p-j}^{-1}\binom{n-p}{q-j}^{-1}\\ &=\binom{n}{p}\binom{n-p}{q}\sum_{k=0}^{\min\{p,q\}}\prod_{j=0}^{k-1}\frac{(j-p)(j-q)}{(j+1)(j+n-p-q+1)}\\ &=\binom{n}{p}\binom{n-p}{q}\sum_{k=0}^{\min\{p,q\}}\frac{(-p)_k(-q)_k}{(n-p-q+1)_k}\,\frac{1}{k!}\\ &=\binom{n}{p}\binom{n-p}{q}{_2F_1}\left(-p,-q;n-p-q+1;1\right)\tag{1}\\ &=\binom{n}{p}\binom{n-p}{q}\frac{\Gamma(n-p-q+1)\Gamma(n+1)}{\Gamma(n-q+1)\Gamma(n-p+1)}\tag{2}\\ &=\binom{n}{p}\binom{n-p}{q}\frac{(n-p-q)!n!}{(n-q)!(n-p)!}\\ &\,\,\color{blue}{=\binom{n}{p}\binom{n}{q}} \end{align*} and the claim follows.

Comment:

• In (1) we write the sum as hypergeometric $_2F_1$ function evaluated at $z=1$ .

• In (2) we recall a theorem from C. F. Gauss [1812] (see e.g. Theorem 2.2.2 in Special Functions by G.E. Andrews, R. Askey and R. Roy) which is \begin{align*} {_2F_1}(a,b;c;1)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}\tag{3} \end{align*} if $\Re(c-a-b)>0$. We derive from (1) $\Re\left((n-p-q+1)-(-p)-(-q)\right)=n+1>0$ and we can apply (3).

Answer 3

First, observe that $$\sum_{k=0}^{\min(p,q)} \begin{pmatrix} n \\ k \end{pmatrix} \begin{pmatrix} n - k \\ p - k \end{pmatrix} \begin{pmatrix} n - p \\ q - k \end{pmatrix} = \sum_{k=0}^{\min(p,q)} \begin{pmatrix} n \\ k \end{pmatrix} \begin{pmatrix} n - k \\ p - k \end{pmatrix} \begin{pmatrix} n - k - (p - k) \\ q - k \end{pmatrix}$$ This means that there is a clear pattern. You could imagine that you have $n$ labeled balls in a box and three buckets: A, B, and C. You take $k$ balls and put them into bucket A. There are $\begin{pmatrix} n \\ k \end{pmatrix}$ ways to do this. Then you take $p - k$ balls from the box and put them in Bucket B. There are $\begin{pmatrix} n - k \\ p - k \end{pmatrix}$ ways to do this. Lastly, you take $q - k$ balls from the box and put them in Bucket C. There are $\begin{pmatrix} n - k - (p - k) \\ q - k \end{pmatrix}$ ways to do this. The total number of ways to execute this procedure is given by the sum on the left-hand side.

Now let's consider a different procedure. Suppose all balls are back in the box and we have stickers with the labels "B" and "C". We choose $p$ balls and label them with the sticker "B". There are $\begin{pmatrix} n \\ p \end{pmatrix}$ ways to do this. Now we choose $q$ balls and label them with the sticker "C". Note that a ball can possibly have two stickers on it, "B" and "C"! There are $\begin{pmatrix} n \\ q \end{pmatrix}$ ways to do this. The total number of ways to apply these stickers is equal to the right-hand side of the equation.

To see that the two values are equal, we now put all the balls with only sticker "B" in bucket B and all the balls with only sticker C in bucket C. We are left with a number of balls that have both stickers, "B" and "C". We put these balls in bucket A. Suppose that $k$ balls ended up in bucket A. Then there are $p - k$ balls in bucket B and $q - k$ balls in bucket C. We see that there is a one-to-one correspondence between the ways of putting the balls into the buckets A, B, and C, and the ways of applying the stickers "B" and "C" to the balls.

As to your diagram, consider drawing a third disjoint blob in the diagram on the right in purple color. Then on the left-hand side, you can create a third region which is the intersection between red and blue, thereby obtaining three disjoint regions.

Answer 4

Since @epi163sqrt has posted an algebraic proof I would like to contribute as well. We seek to show that

$${n\choose p} {n\choose q} = \sum_{k=0}^{\min\{p,q\}} {n\choose k} {n-k\choose p-k} {n-p\choose q-k}.$$

We may extend the sum to infinity with the right coefficient extractors:

$$[z^p] (1+z)^n [w^q] (1+w)^{n-p} \sum_{k\ge 0} {n\choose k} z^k w^k (1+z)^{-k} \\ = [z^p] (1+z)^n [w^q] (1+w)^{n-p} (1+zw/(1+z))^n \\ = [z^p] [w^q] (1+w)^{n-p} (1+z+zw)^n \\ = {n\choose p} [w^q] (1+w)^{n-p} (1+w)^p \\ = {n\choose p} [w^q] (1+w)^n = {n\choose p} {n\choose q}.$$

This is the claim.