Answer to Question #140331 in Discrete Mathematics for Promise Omiponle

Question

Suppose that a department contains 8 men and 20 women. How many ways are there to form a committee with 6 members if it must have strictly more women than men?

Answer 1

Since we have to form a committee of 6 members with strictly more women means that there can be either 4 or 5 or all 6 women in the committee which can be formed in the following number of ways,

₈C₂*₂₀C₄ + ₈C₁*₂₀C₅ + ₈C₀*₂₀C₆

Formula for combination =$\frac {n!}{(n-r)!*r!}$

Substituting the values of n & r in the above formula we get,

₈C₂*₂₀C₄ + ₈C₁*₂₀C₅ + ₈C₀*₂₀C₆ =298452

Hence we can form a committee of 6 members in 298452 ways if it must have strictly more women than men