You are simply choosing 3 people from 8
that would be 8CHOOSE3 or C(8,3)
= 8!/(3!5!)
= 56
the class president plans to randomly select a committee of three people from three boys and five girls. How many committees are possible?
4 answers
thanks
Since it is a combination and order does not matter. You would have 8 students taken 3 at a time.
MATH:
8C3= 8×7×6/3×2×1
=56 possible committees
MATH:
8C3= 8×7×6/3×2×1
=56 possible committees
By the way, the C in 8C3 does NOT stand for CHOOSE. It stands for combination.