# Thread: Simplify the ratio

1. ## Simplify the ratio

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.

2. Originally Posted by flexus
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?

3. 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.

4. Originally Posted by Kasper
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?
thank you i see what you did. n+4