A committee of 5 is to be formed from 6 men and 4 women with at least 3 women. How many ways?

Step 1: Understand the problem constraints for committee formation

We need to form a committee of 5 people from a group of 6 men and 4 women. The primary constraint is that the committee must have 'at least 3 women'. This means we can have either exactly 3 women or exactly 4 women.

Step 2: Case 1: Exactly 3 women in the committee

If there are exactly 3 women in the committee of 5, then the remaining $5 - 3 = 2$ members must be men. We calculate the number of ways to select 3 women from 4 women using combinations, and the number of ways to select 2 men from 6 men using combinations. The total ways for this case is the product of these two combinations.

Step 3: Calculate combinations for Case 1

We use the combination formula $C(n, r) = \frac{n!}{r!(n-r)!}$ to calculate the number of ways. For women, $C(4, 3)$ gives 4 ways. For men, $C(6, 2)$ gives 15 ways. Multiplying these gives 60 ways for the first case.

Step 4: Case 2: Exactly 4 women in the committee

If there are exactly 4 women in the committee of 5, then the remaining $5 - 4 = 1$ member must be a man. We calculate the number of ways to select 4 women from 4 women using combinations, and the number of ways to select 1 man from 6 men using combinations. The total ways for this case is the product of these two combinations.

Step 5: Calculate combinations for Case 2

Again, we use the combination formula. $C(4, 4)$ gives 1 way (as $0! = 1$). $C(6, 1)$ gives 6 ways. Multiplying these gives 6 ways for the second case.

Step 6: Calculate total number of ways

Since these two cases (exactly 3 women and exactly 4 women) are mutually exclusive and cover all possibilities for 'at least 3 women', we add the number of ways from each case to find the total number of ways to form the committee.

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