Question #3f31e

1 Answer
May 16, 2017

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#