Question

How do you find an integer n such that `((n+7)!)/(n!(n+6)(n+4))`?

Answer 1

See below.

Explanation:

`((n+7)!)/(n!(n+6)(n+4))=((n+1)(n+2)cdots(n+7))/((n+6)(n+4))=`
`=(n+1)(n+2)(n+3)(n+5)(n+7)`

As we can see

`((n+7)!)/(n!(n+6)(n+4))=(n+1)(n+2)(n+3)(n+5)(n+7)` and the division always gives an integer as result. In other words

`((n+7)!)/(n!(n+6)(n+4))` is integer for any `n in NN, n ge 0`