In how many ways can a committee of 5 be formed from 6 men and 4 women so that it has at least 2 women?

"At least" questions have two routes: add up every allowed case, or count everything and subtract the forbidden cases. Use whichever has fewer cases. Here the forbidden cases (0 or 1 woman) are only two, so subtraction is quicker, and doing it both ways is a strong check.

Question: 5-person committee from 6 men and 4 women, at least 2 women.

Subtraction method:

1. All committees: C(10, 5) = 252.
2. 0 women: C(6, 5) = 6.
3. 1 woman: C(4, 1) x C(6, 4) = 4 x 15 = 60.
4. Answer = 252 - 6 - 60 = 186.

Direct method:

• 2 women: C(4,2) x C(6,3) = 6 x 20 = 120
• 3 women: C(4,3) x C(6,2) = 4 x 15 = 60
• 4 women: C(4,4) x C(6,1) = 1 x 6 = 6
• Total = 186.

const C = (n, r) => (r === 0 ? 1 : (C(n, r - 1) * (n - r + 1)) / r);
console.log(C(10, 5) - C(6, 5) - C(4, 1) * C(6, 4)); // Output: 186
console.log(C(4, 2) * C(6, 3) + C(4, 3) * C(6, 2) + C(4, 4) * C(6, 1)); // Output: 186

• Choosing 2 women first and then any 3 from the remaining 8 (C(4,2) x C(8,3) = 336), which double counts.
• Adding when you should multiply: within one case, the choices of men and women multiply.