Question

A company's board of directors wants to form a committee of 3 of its members. There are 6 members to choose from. How many different committees of 3 members could possibly be formed?

Answer 1

Order doesn't mater so use combinations:
6C3 = 6!/ (3! (6-3)! ) = 6! /(3!*3!) = 6*5*4*3! / (3! 3!) = 6*5*4/3*2*1) = 20

Answer 2

Answer:

Total 20 different committees of 3 members could possibly be formed.

Step-by-step explanation:

Given information:

Total number of members = 6

Total number of members who are selected = 3

Total number of ways to select r items from n items is

$^nC_r=\frac{n!}{r!(n-r)!}$

Total number of ways to select 3 members from 6 members is

$^6C_3=\frac{6!}{3!(6-3)!}$

$^6C_3=\frac{6\times 5\times 4\times 3!}{3\times 2\times 1\times (3)!}$

Cancel out the common factors.

$^6C_3=\frac{6\times 5\times 4}{3\times 2\times 1}$

$^6C_3=20$

Therefore total 20 different committees of 3 members could possibly be formed.