What is $\binom{n}{2} \binom{n}{1} + \binom{n}{3} \binom{n}{2} + \ldots + \binom{n}{n-1} \binom{n}{n-2} + \binom{n}{n} \binom{n}{n-1}$?

Question

A bag contains $n$ white and $n$ black balls, all of equal size. Balls are drawn at random. Find the probability that there are both white and black balls in the draw and that the number of white balls is greater than that of black balls by $1$.

The total number of possible draws will be the number of subsets of a set with $2n$ elements (minus the empty set) since we have $2n$ balls, hence $2^{2n} - 1$. The number of draws satisfying the conditions there's both white and black in the draw and also the number of white being greater than that of black by $1$ involves computing the following sum:$$\binom{n}{2} \binom{n}{1} + \binom{n}{3} \binom{n}{2} + \ldots + \binom{n}{n-1} \binom{n}{n-2} + \binom{n}{n} \binom{n}{n-1},$$which I don't know how to do. Could anybody help? Thanks. I've tried playing around with paths on an $n \times n$ grid, also an $2n \times n$ grid to come up with a combinatorial proof, not to any success.

It seems that Vandermonde could help here: How to prove Vandermonde's Identity: $\sum_{k=0}^{n}\binom{R}{k}\binom{M}{n-k}=\binom{R+M}{n}$? Or you could have a look at the post with sums similar to this one, such as: Proof of $\sum_{k=1}^n \binom{n}{k} \binom{n}{n+1-k} = \binom{2n}{n+1}$ via induction, Combinatorial identity $\sum_{k=0}^{n}\binom{n}{k}\binom{n}{k -1} = \binom{2n}{n+1}$ or ... — Martin Sleziak, Jan 09 '22 at 15:53
... or Finding a close formula for $\sum_{r=0}^k {k \choose r}{k \choose r-1}$. In case it is useful, I will also add a link to this post: How to search on this site? — Martin Sleziak, Jan 09 '22 at 15:53
Thank you @MartinSleziak, with your help I was able to solve my problem. — user1013124, Jan 09 '22 at 15:58
I am reading your question a bit differently. If I am randomly deciding number of balls to draw and say I decide to draw $(2i+1)$ balls with probability $1/(2n+1)$, the probability that I will draw one more white is ${n \choose i+1}{n \choose i} / {2n \choose 2i+1}$... the way I am looking at it, the denominator is not same for every draw. May be I am interpreting it incorrectly... — Math Lover, Jan 09 '22 at 16:09
@MathLover The confounding part here is the underlying probability distribution. You're assuming first that the number of balls is determined, then the draw is made. However, my interpretation is "every ball is picked with probability 0.5". EG In your distribution, 0 balls are picked with probability $1/(2n+1)$, whereas in mine, 0 balls are picked with probability $1/2^n$. $\quad$ I'm inclined to go with my distribution. — Calvin Lin, Jan 09 '22 at 16:13
@CalvinLin I see now. OP's / your interpretation is the right interpretation (at least the most likely interpretation). Also I think you meant zero balls are drawn with probability $1/2^{2n}$ — Math Lover, Jan 09 '22 at 16:22

score 3 · Accepted Answer · edited Jan 09 '22 at 20:02

A combinatoric proof : Let : $$A = \{1, 2, \ldots, 2 k\}, B = \{1, \ldots, k\} \text{ and } C = \{k + 1, \ldots, 2 k\}$$ We need to choose $k - 1$ elements from $A$. There are two ways :

First way : Choose $k - 1$ element directly for $A$ : In this case, we have then : $$\binom{2 k}{k - 1}$$ ways.
Second way : We can choose $r$ elements from $B$ and $k - 1 - r$ from $C$ for $r \in \{0, \ldots, k -1 \}$ : In this case we have : $$\sum_{r = 0}^{k-1} \binom{k}{r} \binom{k}{k - 1 - r}$$ ways.

We deduce that : $$\binom{2 k}{k - 1} = \sum_{r = 0}^{k-1} \binom{k}{r} \binom{k}{k - 1 - r} = \sum_{r = 0}^{k-1} \binom{k}{r} \binom{k}{r + 1}$$

What is $\binom{n}{2} \binom{n}{1} + \binom{n}{3} \binom{n}{2} + \ldots + \binom{n}{n-1} \binom{n}{n-2} + \binom{n}{n} \binom{n}{n-1}$?

1 Answers1