A committee of 5 people is to be chosen from a group of 6 men and 4 women. How many committees are possible if there is to be a majority of women?

Analysis

The problem asks us to form a committee of 5 people from a total group of 10 (6 men and 4 women). The condition is that there must be a majority of women.

Concept

Since the committee size is 5, a "majority of women" means there must be more women than men. The possible compositions for the committee are:

1. 3 women and 2 men
2. 4 women and 1 man

We use the combination formula, denoted as $^nC_r = \frac{n!}{r!(n-r)!}$, where $n$ is the total number of items and $r$ is the number of items to be chosen.

Step-by-Step Solution

Case 1: 3 women and 2 men

• Number of ways to choose 3 women from 4: $^4C_3 = \frac{4!}{3!(4-3)!} = 4$
• Number of ways to choose 2 men from 6: $^6C_2 = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15$
• Total for Case 1: $4 \times 15 = 60$

Case 2: 4 women and 1 man

• Number of ways to choose 4 women from 4: $^4C_4 = 1$
• Number of ways to choose 1 man from 6: $^6C_1 = 6$
• Total for Case 2: $1 \times 6 = 6$

Total Committees To get the final answer, add the results of the two cases: $60 + 6 = 66$

Conclusion

The total number of possible committees with a majority of women is 66.

Exam Tip

When you see "majority" in committee problems, always list out the possible scenarios first. If the committee size is odd (like 5), you don't have to worry about a tie, but always check if the number of available people in each category is enough to satisfy the condition (e.g., you couldn't have 5 women because there are only 4 available).