Solutions to $\binom{n}{5} = 2 \binom{m}{5}$ In Finite Mathematics by Lial et al. (10th ed.), problem 8.3.34 says:

On National Public Radio, the Weekend Edition program 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 $(m + 2)(m  +1) = 2(m - 3)(m - 4)$.  I have ruled out $n = m + 3$ or $n = m + 4$ with similar checks.  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 is now a question at MathOverflow.
 A: Many Diophantine equations are solved using modern algebraic geometry. For an informal survey how this works, see
M. Stoll, How to solve a Diophantine equation, arXiv.
The most prominent example is Fermat's equation. But there are also interesting binomial equations. It has been shown very recently that $\binom{x}{5}=\binom{y}{2}$ has exactly $20$ integer solutions:
Y. Bugeaud, M. Mignotte, S. Siksek, M. Stoll, Sz. Tengely, Integral Points on Hyperelliptic Curves, arXiv
I don't know if this can be proven by elementary means. And I don't know the situation for $\binom{x}{5}=2 \binom{y}{5}$. I just want to warn you that it might be a waste of time to look for elementary solutions, and that instead more sophisticated methods are necessary. On the other hand, this equation arises as a problem from a book, so I am not sure ...
A: WRONG The problem generalizes to the drawing of $N>1$ balls.  The solution for any $N$ is characterized by a telescoping product of probabilities (one for each ball drawn) whose first factor is $\frac{2N-1}{2N}$ with subsequent factors progressively smaller.  The solutions are unique UNPROVEN: consider perturbations.  Increasing (or decreasing) the first factor increases (or decreases WRONG) all the factors and so the product.  Thus the first factor of any solution is fixed.  With the first factor fixed, some measure of the overall shrinkage of subsequent factors must be constant if the product is not to change; but if we multiply the numerator and denominator of the first factor by some natural number greater than 1, we reduce the relative shrinkage from factor to factor, which increases the product.  Thus the telescoping product is immutable; so every solution has $2N$ balls with all but one being blue.  [Apparently rude edits added by author.]
