How would I solve a combination problem where one combination equals 5 times another combination.

Question

I have $ \frac{n!}{4!(n-4)!} = 5\frac{n!}{6!(n - 6)!} $

and $n \geq 6$

I tried putting the equation into wolfram alpha and receive a solution, but couldn't see a step by step guide on how to perform the manipulation correctly.

Answer 1

Start with $$\frac{n!}{4!(n-4)!}=5\frac{n!}{6!(n-6)!}\\\frac{n!}{n!}=\frac{4!\cdot 5}{6!}\frac{(n-4)!}{(n-6)!}$$ and keep going: there should be a lot of cancellation using the factorials. Below is a spoiler which you can use to double check your solution (it is a continuation of the above computation).

$$\\1=\frac{1}{6}(n-4)(n-5)\\6=n^2-9n+20\\n^2-9n+14=0\\(n-7)(n-2)=0$$