Respuesta :
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
6C3 = 6!/ (3! (6-3)! ) = 6! /(3!*3!) = 6*5*4*3! / (3! 3!) = 6*5*4/3*2*1) = 20
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
[tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]
Total number of ways to select 3 members from 6 members is
[tex]^6C_3=\frac{6!}{3!(6-3)!}[/tex]
[tex]^6C_3=\frac{6\times 5\times 4\times 3!}{3\times 2\times 1\times (3)!}[/tex]
Cancel out the common factors.
[tex]^6C_3=\frac{6\times 5\times 4}{3\times 2\times 1}[/tex]
[tex]^6C_3=20[/tex]
Therefore total 20 different committees of 3 members could possibly be formed.