Given that $\frac{n!}{(n-x)!}+\frac{x!}{(x-n)!}=48$, find the value of $\binom{n+4}{x-3}$

Question

I am struggling with the following:

Given that $\frac{n!}{(n-x)!}+\frac{x!}{(x-n)!}=48$, find the value of $~^{n+4}C_{x-3}$.

Any help with the method would be great. The answer is 8.

Answer 1

In order for this to make sense both $n-x$ and $x-n$ must be nonnegative, which implies that $x = n$. Thus $2n! = 48 \implies x = n = 4$.

Answer 2

HINT:

Case$\#1:$ What if $n=x$

Case$\#2:$ If $n>x,$ $$\dfrac1{(x-n)!}=0$$

$$\dfrac{n!}{(n-x)!}=n(n-1)\cdots \{n-(x+1)\}$$

Check with $n-x=1,2,3,4$ as for $n-x=5, 5!>48$

Case$\#3:$ Similarly if $n<x$