simplify the ratio
(n+4)! / (n+3)!
ok i know how to simplify ratios like 4!/3! which = 4 but when they put the n in the problem i dont know what to do.
Just like $\displaystyle 5!=(5)(4)(3)(2)(1)$, we can use the same rationale here;
$\displaystyle (n+4)!=(n+4)(n+4-1)(n+4-2)...$
$\displaystyle \frac{(n+4)!}{(n+3)!} = \frac{(n+4)(n+3)(n+2)(n+1)...}{(n+3)(n+2)(n+1)...}$
The multiplications cancel as the terms lower to the right, can you see what might be left over?
Remember what a factorial is and write it out, then cancel.
Like so:
$\displaystyle (n+4)!=(n+4)(n+3)(n+2)(n+1)(n)............$
$\displaystyle (n+3)!=(n+3)(n+2)(n+1)(n).................$
So, $\displaystyle \frac{(n+4)!}{(n+3)!}=\frac{(n+4)(n+3)(n+2)(n+1)(n )............}{(n+3)(n+2)(n+1)(n)................. }$
See the cancellations?. You are left with n+4.