How can I simplify (n+3)!/(n+1)! using the property of factorials?

[Blade]

How would I go about simplifying something like:
(n+3)!/(n+1)! ?

 

[cookiemonster]

[tex]\frac{(n+3)!}{(n+1)!} = \frac{(n+3)(n+2)\cdot(n+1)!}{(n+1)!}=(n+2)(n+3) = n^2 + 5n + 6[/tex]

cookiemonster

 

[M98Ranger]

To simplify this expression, you can use the property of factorials that states: n!/(n-k)! = n(n-1)(n-2)...(n-k+1). In this case, n = n+3 and k = n+1, so the expression becomes:

(n+3)!/(n+1)! = (n+3)(n+2)(n+1)/[(n+1)n(n-1)...(3)]

The (n+1) terms in the numerator and denominator cancel out, leaving us with:

(n+3)!/(n+1)! = (n+3)(n+2)/n(n-1)...(3)

This is the simplified expression for (n+3)!/(n+1)!.