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 $(m + 2)(m  +1) = 2(m - 3)(m - 4)$.  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.
 A: This isn’t a complete answer, but the problem “reduces” to finding the finitely many rational points on a certain genus 2 hyperelliptic curve. This is often possible by a technique
involving a reduction to finding the rational points on a finite set of rank 0 elliptic curves—see for example “Towers of 2-covers of hyperelliptic curves” by Bruin and Flynn in Trans. Amer. Math. Soc. 357 (2005) #11 4329–4347.
In your case, the curve is $u^2= 9t^6+16t^5-200t^3+256t+144$. There are the following 16
rational points: $(t,u)$ or $(t,-u)= (0,12)$, $(1,15)$, $(2,12)$, $(4,204)$, $(-1,9)$, $(-2,36)$, $(-4,180)$, and $(7/4,411/64)$. If these are the only rational points then the only non-trivial solution to your equation is $n=10$, $m=9$. To see this suppose that $$n(n-1)(n-2)(n-3)(n-4)=2m(m-1)(m-2)(m-3)(m-4).$$ Let $$y=(n-2)^2\text{ and }x=(m-2)^2.$$ Squaring both sides we find that $$y(y-1)(y-1)(y-4)(y-4)=4x(x-1)(x-1)(x-4)(x-4).$$ Suppose $y$ isn’t $0$. Then $4x/y$ is $t^2$ for some rational $t$ with $$(y-1)(y-4)=t(x-1)(x-4).$$ We replace $y$ by $4x/t^2$ in this equation and find that $$(t^5-16)x^2-(5t^5-20t^2)x+(4t^5-4t^4)=0.$$ So this last quadratic polynomial in $x$ has a rational root and its
discriminant is a square. This gives the hyperelliptic curve above. Note that the case $n=10$, $m=9$ of your problem corresponds to the point $(7/4,411/64)$ on this curve.
EDIT:
More generally one can look for rational $m$ and $n$ with $[n]_5= 2[m]_5$. If $(t,u)$ is a
rational point on the hyperelliptic curve with $t$ non-zero, set $x=(5t^5-20t^2+t^2u)/(2t^5-32)$ and $y=4x/t^2$. Then if $x$ is a square in $\mathbb Q$, one gets such an $m$ and $n$ with $m=2+\sqrt x$ and $n=2+\sqrt y$. Joro’s points lead in this way to the solutions
$(n,m)=(10,9)$, $(10/3,5/3)$, and $(78/23,36/23)$, the last one being rather unexpected. (And as François notes, each $(n,m)$ gives a second solution $(4-n,4-m)$.) Perhaps these solutions and
the trivial solutions with $m$ and $n \in \{0,1,2,3,4\}$ are the only rational solutions.
