In how many ways can a committee of 3 persons be chosen out of 8 persons?

Question

Accepted Answer

The number of ways to choose 3 persons out of 8 is given by the combination $^8C3 = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} = \frac{8 	imes 7 	imes 6}{3 	imes 2 	imes 1} = 56$. Therefore, option C is correct.

Permutations and Combinations

❓ Sibling Questions in this Chapter