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

*print*Print*list*Cite

### 2 Answers

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!).`

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!