From a group of 6 men and 4 women, in how many different ways can a committee of 5 persons be formed so that at least 2 members of the committee are women?

Introduction / Context:
This question is a standard combinations problem with an additional constraint. You need to form a committee of 5 members from a mix of men and women, with the condition that at least 2 women must be included. It tests the ability to handle combination formulas and to sum over valid cases that satisfy the constraint.

Given Data / Assumptions:

• Total men available = 6.
• Total women available = 4.
• Committee size = 5 persons.
• At least 2 women must be included.
• Every person is distinct, and each combination of individuals counts as a different committee.

Concept / Approach:
We need to consider all valid distributions of women and men that satisfy the condition "at least 2 women." The possible cases for the committee of 5 are:

• Case 1: 2 women and 3 men.
• Case 2: 3 women and 2 men.
• Case 3: 4 women and 1 man.

For each case, we use combinations: Number of ways = C(4, number of women) * C(6, number of men). Finally, we sum the results of all valid cases.

Step-by-Step Solution:
Step 1: Case 1 (2 women, 3 men) Ways to choose women = C(4,2). Ways to choose men = C(6,3). Total for Case 1 = C(4,2) * C(6,3) = 6 * 20 = 120. Step 2: Case 2 (3 women, 2 men) Ways to choose women = C(4,3) = 4. Ways to choose men = C(6,2) = 15. Total for Case 2 = 4 * 15 = 60. Step 3: Case 3 (4 women, 1 man) Ways to choose women = C(4,4) = 1. Ways to choose men = C(6,1) = 6. Total for Case 3 = 1 * 6 = 6. Step 4: Add all valid cases: 120 + 60 + 6 = 186. Step 5: Therefore, total number of committees with at least 2 women = 186.

Verification / Alternative check:
As a check, we can compute the total number of committees without restriction and subtract those with fewer than 2 women. Total committees from 10 people = C(10,5) = 252. Committees with 0 women: choose 5 men from 6 = C(6,5) = 6. Committees with 1 woman: choose 1 woman from 4 and 4 men from 6 = C(4,1) * C(6,4) = 4 * 15 = 60. Committees with fewer than 2 women = 6 + 60 = 66. Committees with at least 2 women = 252 - 66 = 186, which matches our earlier answer.

Why Other Options Are Wrong:
196 and 190: These are close but result from small counting mistakes, such as miscomputing one of the combination terms. 200: A rounded guess that does not match any careful combination calculation. 176: This undercounts and likely omits some valid case or miscalculates C(6,3) or similar terms.

Common Pitfalls:
Students sometimes consider only one distribution, such as exactly 2 women, instead of all cases with at least 2 women. Another error is to forget that combinations, not permutations, should be used because order within the committee does not matter. Some also double count by mixing cases or forget to include the case with all 4 women. Listing the cases systematically and using correct combination formulas helps avoid these issues.

Final Answer:
The number of possible committees of 5 persons with at least 2 women is 186.