# 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 $2 \binom{n}{5}$.

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The equation defines a smooth quintic so there are only finitely many solutions. –  Felipe Voloch Apr 19 '13 at 3:17
Any solutions have $n=\lfloor 2^{1/5} m\rfloor$, so there are no further solutions for $m\le 10,000,000$. –  Brendan McKay Apr 19 '13 at 4:43
For what it's worth: I've checked that there are no other solutions for n up to 35 billion or so. –  shreevatsa Apr 19 '13 at 5:01

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= 9*t^6+16*t^5-200*t^3+256*t+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 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^5-20*t^2+t^2*u)/(2*t^5- 32) and y=4x/t^2. Then if x is a square in Q, one gets such an m and n with m=2+(a square root of x) and n=2+(a square root of 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 Francois 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.

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Looks like there are more points on the curve. $t$ are: [-4, -13/5, -2, -1, -5/8, -1/2, -8/27, 0, 26/41, 1, 7/4, 2, 36/17, 4] –  joro Apr 19 '13 at 12:00
According to Magma the Jacobian $J$ of this hyperelliptic curve has rank 4 and satisfies End(J)=Z over the complex numbers ; in particular this genus 2 curve has no elliptic quotient. –  François Brunault Apr 19 '13 at 12:20
The curve C defined by $\begin{pmatrix} n \\ 5\end{pmatrix} =2 \begin{pmatrix}m \\ 5\end{pmatrix}$ has genus 6. I guess the genus 2 curve in your answer is the quotient of C by the involution $(m,n) \to (4−m,4−n)$, which is the only obvious automorphism. –  François Brunault Apr 19 '13 at 12:55
Using Stahlke and Stoll's program ratpoints, I find that there are no further rational solutions of $u^2 = 9t^6+16t^5-200t^3+256t+144$ with $u=r/s$ and $|r|,|s| \leq 10^5$. On the original equation, it should be easy to show there are no further solutions up to say $10^{100}$ using the fact that $(n-2)/(m-2) = 2^{1/5} + O(1/m)$, so there are only a few hundred approximations to $2^{1/5}$ that are close enough, and they can be computed via the continued-fraction expansion. –  Noam D. Elkies Apr 22 '13 at 4:45