A Committee with At Least Two Men

“At least two” can be handled by adding the qualifying cases or by the complement. Here the qualifying cases (2, 3 or 4 men) are few enough to add directly, using the multiplication and addition principles.

A committee of $4$ is chosen from $5$ men and $4$ women. How many committees contain at least $2$ men? [5]

The working

“At least $2$ men” means $2$, $3$, or $4$ men (with the rest women). Count each case (men $\times$ women), then add.

Case 1, $2$ men and $2$ women: $^{5}C_{2} \times {}^{4}C_{2} = 10 \times 6 = 60 \quad \text{(M1)}$

Case 2, $3$ men and $1$ woman: $^{5}C_{3} \times {}^{4}C_{1} = 10 \times 4 = 40 \quad \text{(M1)}$

Case 3, $4$ men and $0$ women: $^{5}C_{4} \times {}^{4}C_{0} = 5 \times 1 = 5 \quad \text{(A1)}$

Add the cases (“or” between them): $60 + 40 + 5 = 105 \quad \text{(M1, A1)}$

Where the marks are won and lost

• List all qualifying cases: $2$, $3$ and $4$ men. Missing the $4$-men case (or the $0$-women $^{4}C_{0} = 1$) undercounts.
• Within each case, multiply (men and women choices are independent); add across cases (“at least 2” is one case or another).
• $^{4}C_{0} = 1$ (one way to choose nobody), don’t drop the all-men case.

Common mistakes

• Forgetting the $4$-men (all-men) case.
• Adding within a case or multiplying across cases (operations reversed).
• Using permutations instead of combinations (a committee is unordered).

Full method: Combinations (nCr) notes. See also At Least One Selection. Topic home: Permutations & Combinations pillar.