Question #3f31e

1 Answer
May 16, 2017

Answer: 2121

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