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joja [24]
3 years ago
7

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 diffe

rent committees of 3 members could possibly be formed?
Mathematics
2 answers:
kondaur [170]3 years ago
5 0
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
Nata [24]3 years ago
5 0

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.

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