Solving equations involving permutations and combinations

Question

I need to solve for $n$ if $$P^n_4=60\times C^n_2$$I’ve gotten up to $$n(n-1)(n-2)(n-3)=30n(n-1)$$

Answer 1

Now your equation can be simplified to

$n(n-1)[(n-2)(n-3)-30]=0$ which can be further simplified to

$n(n-1)[n^2-5n-24]=0$

From here we get following values

$n=-3,0,1,8$

Now since $n\geq4$

$n=8$ is your answer.

Answer 2

By eliminating common terms on both sides, you get $(n-2)(n-3)=30$, or $n^2-5n-24=0$. Positive solution is $n=8$.

Answer 3

You are nearly there! You got$$n(n-1)(n-2)(n-3)=30n(n-1)$$Now shift all the terms to one side to get$$n(n-1)\Big[(n-2)(n-3)-30\Big]=0$$So $n$ could be $0,1$ or such that$$(n-2)(n-3)-30=0\\\implies n^2-5n-24=(n-8)(n+3)=0\\\therefore n=0,1,-3,8$$If you define $P^n_4=\displaystyle\frac{n!}{(n-4)!}$, you need $n\ge4$. This leaves you with the answer $n=8$.