Answer to Question #140242 in Statistics and Probability for Mawyella

Question

From a group of 10 men and 15 women, how many committees of size 9 are possible a) with no restrictions; b) with 6 men and 3 women; c) with 5 men and 4 women if a certain man must be on the committee?

Answer 1

Given that,

There are 10 men and 15 women

Answer a)

Since there are no restrictions for the total number of persons in the committee, the number of possible combinations is given by,

C = $\frac {n!}{(n-r)!*r!}$

where,

n = 10 men + 15 women = 25

r = 9

Hence, ₂₅C₉ = $\frac {25!}{(25-9)!*9!}=\ 2,042,975$ ways

Answer b)

For a committee consisting of 6 men and 3 women the number of combinations = 10C6*15C3 = 95,550 ways

Answer c)

Given that a certain man out of 5 men should be in the committee then the number of remaining people can selected in following number of ways

9C4*15C4 = 171,990 ways