Question
In how many ways can a committee of 3 men and 2 women be selected from a group of 6 men and 5 women?
Solution — Step by Step
We are choosing (selecting) people for a committee — order doesn’t matter (a committee of {A, B, C} is the same regardless of who is listed first). This is a combination problem.
We need to select:
- 3 men from 6 (one selection)
- 2 women from 5 (another selection)
Since these are independent selections (choosing men doesn’t affect choosing women), we multiply the results.
Since we must choose men AND women (both selections), we multiply:
Why This Works
The Multiplication Principle (or Fundamental Counting Principle) states: if task A can be done in ways and task B can be done in ways, and the two tasks are independent, then both can be done together in ways.
Selecting 3 men and selecting 2 women are independent — they involve different people from different groups. For each of the 20 ways to choose the men, there are 10 ways to choose the women, giving total committees.
Alternative Method — Cross-Verification
Total people = 11. Total ways to choose any 5 from 11 = . Our answer of 200 (committees with exactly 3 men and 2 women) is less than 462 (committees of any composition), which makes sense as a sanity check.
For problems asking “at least” or “at most” a certain number of one gender: list all valid compositions and add. For example, “at least 2 men from 6M, 5W, selecting 5 total” = (2M, 3W) + (3M, 2W) + (4M, 1W) + (5M, 0W). Calculate each and sum.
Common Mistake
Students sometimes use permutations () instead of combinations () for committee selection. Committees are unordered — {Alice, Bob, Carol} is the same committee as {Bob, Carol, Alice}. If the problem asked for an “arranged committee” or “committee where positions matter” (e.g., a team with a captain), then permutations would apply. When in doubt: if order matters, use ; if order doesn’t matter, use .