Question

Question #3f31e

Answer 1

Answer: `21`

Explanation:

Evaluate `((7),(5))`

This is a combination, in general form:
`((n),(k))=(n!)/(k!(n-k)!)`

So, for this problem, we have:
`((7),(5))`
`=(7!)/(5!(7-5)!)`
`=(7!)/(5!(2!))`

Note that we can re-write `7!` as `7*6*5!` to cancel with the `5!` in the denominator:
`=(7*6*5!)/(5!(2!))`
`=(7*6)/(2)`
`=7*3`
`=21`