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.
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1Try to cancel some terms from both sides, using the definition of factorial. You will certainly get a more simplified equation. – Sarvesh Ravichandran Iyer May 10 '18 at 02:19
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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).
$$\\1=\frac{1}{6}(n-4)(n-5)\\6=n^2-9n+20\\n^2-9n+14=0\\(n-7)(n-2)=0$$
Dave
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How would I properly cancel $\frac{(n-4)}{(n-6)!}$ to get $(n-4)(n-5)$ I take it that $\frac{(n-4)(n-3)(n-2)(n-1)n}{(n-6)(n-5)(n-4)(n-3)(n-2)(n-1)n}$ canceling down to $(n-6)(n-5)$ is not correct. – ZeroPhase May 10 '18 at 02:46
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This is incorrect. We have $$(n-4)!=(n-4)(n-5)(n-6)\cdots(2)(1)$$ and $$(n-6)!=(n-6)(n-7)(n-8)\cdots (2)(1)$$ – Dave May 10 '18 at 02:47