Combinations: Selecting a Committee

Committee questions are combinations (order within the committee does not matter) combined by the multiplication principle. “Exactly 2 men and 2 women” means choose the men, choose the women, and multiply, because the two choices are independent.

A committee of $4$ people is to be chosen from $6$ men and $5$ women. In how many ways can the committee be chosen if it must contain exactly $2$ men and $2$ women? [4]

The working

Step 1, choose the men. Order does not matter, so use combinations, $2$ from $6$: $^{6}C_{2} = \frac{6 \times 5}{2 \times 1} = 15 \quad \text{(M1)}$

Step 2, choose the women, $2$ from $5$: $^{5}C_{2} = \frac{5 \times 4}{2 \times 1} = 10 \quad \text{(M1)}$

Step 3, multiply (independent stages, ”$2$ men and $2$ women”): $^{6}C_{2} \times {}^{5}C_{2} = 15 \times 10 = 150 \quad \text{(M1, A1)}$

Where the marks are won and lost

• Combinations, not permutations: a committee is an unordered group, so $^{6}C_{2}$, not $^{6}P_{2}$.
• “And” means multiply. The men-choice and women-choice are independent, so the counts multiply. Adding them ($15 + 10 = 25$) answers a different question.
• Compute each $^nC_r$ carefully: $^{6}C_{2} = 15$ and $^{5}C_{2} = 10$ (not $30$ and $20$, which would be $^nP_r$).

Common mistakes

• Adding the two combinations instead of multiplying.
• Using $^{6}P_{2}$ / $^{5}P_{2}$ (order does not matter within the committee).
• Choosing $4$ from all $11$ people ($^{11}C_{4}$) and ignoring the “exactly 2 and 2” restriction.

Full method: Combinations (nCr) notes. See also Arrangements & Selections. Topic home: Permutations & Combinations pillar.