# Combinatorics: Choosing Committees

Here's the question that I came across today: There are 10 men and 12 women in a group. A committee of 5 people from the group is to be chosen. How many committees containing least two women are possible? I managed to get the correct answer to the question by computing $C^{22}_{5} - C^{10}_{5} - 12C^{10}_{4}$.

However, why does $C^{12}_{2} \times C^{20}_{3}$ not give the same result? Since the first part of the product covers the number of ways of choosing two women, and the second part gives all the possible ways of choosing the remaining members of the committee.

$^{12}C_2\,^{20}C_3$ counts the ways to select two women as heads of the committee and 3 of the remaining members as the rest of the committee. However, you don't want those two women to hold special positions from any other, we mustn't select from the women twice.
Instead, your answer, via complements, is correct.   You have counted the ways not to form a committee with less than two women.   That is indeed: $~{^{22}C_5} - {^{10}C_5}-12{^{10}C_4}$
Consider a simpler example of having to select a committee of three people with at least two women, from three women ($\rm A,B,C$) and one man ($\rm X$).
${^3C_2}\;{^2C_1}= 6$ would count ways to select two from three women then one from the two remaining.   That is: $\rm \{A,B,C\}, \{A,B,X\}, \{A,C,B\}, \{A,C,X\}, \{B,C,A\}, \{B,C,X\}$. Opps; several sets are just reordered, but order is not important; there are no privileged positions.
We should instead count ${^3C_2}\cdot 1 +{^3C_3} = 4$; ways select two women and one man, or select three women.   That is: $\rm \{A,B,X\}, \{A,C,X\}, \{B,C,X\}, \{A,B,C\}$ Okay!   Every distinct set is counted just once each.