Modern Math (P&C Practice Test 2026 | Exam Prep

A committee of 5 people is to be chosen from a group of 8 men and 7 women. If the committee must consist of exactly 3 men and 2 women, how many different committees can be formed?

tests the fundamental principle of counting, specifically combinations. We need to select 3 men from 8, which can be done in $\binom{8}{3}$ ways. We also need to select 2 women from 7, which can be done in $\binom{7}{2}$ ways. Since these selections are independent, we multiply the number of ways to get the total number of committees. $\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$ $\binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21$ Total committees = $\binom{8}{3} \times \binom{7}{2} = 56 \times 21 = 1176$. Wait, let me recheck my calculation. $\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$. $\binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21$. $56 \times 21 = 1176$. Ah, there seems to be a mistake in my initial options. Let me re-calculate the options.

requires the application of the combination formula, as the order in which committee members are chosen does not matter. We need to select 3 men out of 8, which can be done in $\binom{8}{3}$ ways. We also need to select 2 women out of 7, which can be done in $\binom{7}{2}$ ways. By the multiplication principle, the total number of ways to form such a committee is the product of these two combinations. $\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$. $\binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21$. Total committees = $\binom{8}{3} \times \binom{7}{2} = 56 \times 21 = 1176$. Option B is likely the result of incorrectly calculating permutations or summing instead of multiplying. Option C might arise from misinterpreting the numbers or using a flawed formul