In *Finite Mathematics* by Lial et al. (10th ed.), problem 8.3.34 says:

On National Public Radio, the

Weekend Editionprogram posed the following probability problem: Given a certain number of balls, of which some are blue, pick 5 at random. The probability that all 5 are blue is 1/2. Determine the original number of balls and decide how many were blue.

If there are $n$ balls, of which $m$ are blue, then the probability that 5 randomly chosen balls are all blue is $\binom{m}{5} / \binom{n}{5}$. We want this to be $1/2$, so $\binom{n}{5} = 2\binom{m}{5}$; equivalently, $n(n-1)(n-2)(n-3)(n-4) = 2 m(m-1)(m-2)(m-3)(m-4)$. I'll denote these quantities as $[n]_5$ and $2 [m]_5$ (this is a notation for the so-called "falling factorial.")

A little fooling around will show that $[m+1]_5 = \frac{m+1}{m-4}[m]_5$. Solving $\frac{m+1}{m-4} = 2$ shows that the only solution with $n = m + 1$ has $m = 9$, $n = 10$.

**Is this the only solution?**

You can check that $n = m + 2$ doesn't yield any integer solutions, by using the quadratic formula to solve $(m + 2)(m +1) = 2(m - 3)(m - 4)$. For $n = m + 3$ or $n = m + 4$, I have done similar checks, and there are no integer solutions. For $n \geq m + 5$, solutions would satisfy a quintic equation, which of course has no general formula to find solutions.

Note that, as $n$ gets bigger, the ratio of successive values of $\binom{n}{5}$ gets smaller; $\binom{n+1}{5} = \frac{n+1}{n-4}\binom{n}{5}$ and $\frac{n+1}{n-4}$ is less than 2—in fact, it approaches 1. So it seems possible that, for some $k$, $\binom{n+k}{5}$ could be $2 \binom{n}{5}$.

This question was previously asked at Math StackExchange, without any answer, but some interesting comments were made. It was suggested that I ask here.