# rewrite the following using fatorial notation (show working) (n+5)(n+4)(n+3)(n+2)(n+1)

ahalperi | Certified Educator

Because  (n+5)! = (n+5)(n+4)(n+3)(n+2)(n+1)n!  (by definition)

If we choose to multiply by one in the form of   n!/n! then

(n+5)(n+4)(n+3)(n+2)(n+1)     =

n!

(n+5)(n+4)(n+3)(n+2)(n+1) •    _____    =

n!

(n+5)(n+4)(n+3)(n+2)(n+1)n!           (n+5)!
_________________________  =   _______

n!                                       n!

sciencesolve | Certified Educator

You need to multiply and divide `(n+5)(n+4)(n+3)(n+2)(n+1)` by the product `n*(n-1)*(n-2)*...*3*2*1` such that:

`(n+5)(n+4)(n+3)(n+2)(n+1) = ((n+5)(n+4)(n+3)(n+2)(n+1)*n*(n-1)*(n-2)*...*3*2*1)/(n*(n-1)*(n-2)*...*3*2*1)`

You need to remember that `n*(n-1)*(n-2)*...*3*2*1 = n!` and `1*2*3*...*n*(n+1)*(n+2)*(n+3)*(n+4)*(n+5) = (n+5)!`

`(n+5)(n+4)(n+3)(n+2)(n+1) = ((n+5)!)/(n!)`

Hence, you may write `(n+5)(n+4)(n+3)(n+2)(n+1)` using factorial notation, such that `(n+5)(n+4)(n+3)(n+2)(n+1) = ((n+5)!)/(n!).`