A number theory problem by Megha Malyala

$\begin{aligned} \frac {\frac {n!}{2!(n-2)!}}{\frac {n!}{4!(n-4)!}} & = \frac 21 \\ \frac {4!(n-4)!}{2!(n-2)!} & = 2 \\ \frac {3\times 4}{(n-2)(n-3)} & = 2 \\ (n-2)(n-3) & = 6 \\ n^2 - 5n + 6 & = 6 \\ n(n-5) & = 0 \\ \implies n & = \boxed{5} & \small \color{#3D99F6} n = 0 \text{ is unacceptable.} \end{aligned}$

Since $\frac{n!}{4!(n-4)!}$ is $\frac{1}{2}$ of $\frac{n!}{2!(n-2)!}$ , $\frac{n!}{2!(n-2)!} = 2 \frac{(n!)}{4!(n-4)!}$ . After we cross multiply and simplify the factorials, we now have $(24)(n)!(n-4)! = 4(n)!(n-2)!)$ .

After dividing by $n!$ , we get $(24)(n-4)! = 4(n-2)!)$ . From this, $\frac{(n-2)!}{(n-4)!} = 6$ . This is the critical step - when we expand $(n-2)!$ , we have $(n-2)(n-3)(n-4) \cdots$ , which becomes $(n-2)(n-3)(n-4)!$ . Now, our previous expression becomes $(n-2)(n-3) = 6$ . Now we have a quadratic expression: $(n^2-5n+6) - 6 = 0$ . We can simplify this into $n(n-5) = 0$ .

From this, either $n=0$ , which is possible, but nonsensical since $0:0$ can be in any ratio you want. The remaining solution is $n-5 = 0$ , which makes $n=5$ . We can verify this solution by checking the LHS, which is $\frac{5!}{2!3!} = \frac{5*4}{2} = 10$ ; and the RHS, which is equal to $\frac{5!}{4!} = 5$ . Therefore, the ratio of LHS:RHS is $10:5$ , which is equal to $2:1$ . This verifies the condition in the statement, and therefore our answer.