How to Solve Committee Selection Combination Problems
Solve committee selection aptitude problems — required members, exclusions, subgroup quotas — with worked examples and practice questions.
Expected Interview Answer
Committee selection problems are pure combinations — nCr — because a committee is defined by which people are in it, not by any internal order, and constrained versions (specific people must/must not be included, or subgroups must be represented) are solved by splitting the selection into independent stages and multiplying or summing the cases.
The base case, choosing r people from n with no restriction, is nCr = n!/(r!(n−r)!). When a specific person must be included, fix that seat and choose the remaining r−1 from the other n−1, giving (n−1)C(r−1). When a specific person must be excluded, choose all r from the remaining n−1, giving (n−1)Cr. When the committee must include a fixed number from each of several subgroups (e.g., 2 men and 3 women), compute each subgroup’s combination separately and multiply them together via the multiplication principle. For “at least” conditions, it is often fastest to sum the valid cases or use the complement: total minus the cases that violate the condition.
- One base formula, nCr, anchors every committee variant
- Fixing required/excluded members reduces the problem to a smaller nCr
- Subgroup quotas multiply independent combination counts together
AI Mentor Explanation
Selecting 11 players for a match from a 15-man squad, where the star all-rounder must be included, is solved by fixing that one player’s spot and choosing the remaining 10 from the other 14 players: 14C10. If a specific injured player must be excluded instead, the selection becomes choosing all 11 from the remaining 14 fit players: 14C11. Either way, the committee-selection technique is to lock in the forced condition first, then apply nCr to whatever remains, never re-adding order since the final XI is just a group.
Worked example (with a required member)
Fix required member
- X is always in the committee
Choose the rest
- Pick 3 from remaining 9
- 9C3 = 84
Total committees
- 84 valid committees
Step-by-Step Explanation
Step 1
Confirm order does not matter
A committee is a group; use nCr, not nPr.
Step 2
Handle forced inclusion/exclusion
Fix required members and choose the rest from the reduced pool.
Step 3
Handle subgroup quotas
Compute each subgroup’s combination separately, then multiply them.
Step 4
Handle “at least” conditions
Sum the valid cases, or subtract the invalid cases from the unrestricted total.
What Interviewer Expects
- Recognizing that committee membership is unordered, so nCr applies
- Correctly reducing the pool when a member must be included or excluded
- Multiplying independent subgroup combinations for quota-based selections
- Correctly applying the complement or summation for “at least” conditions
Common Mistakes
- Using nPr instead of nCr because the problem sounds like it involves “choosing roles”
- Forgetting to reduce both n and r when a required member is fixed
- Adding subgroup combinations instead of multiplying them
- Double-counting or under-counting in “at least” scenarios by mishandling the complement
Best Answer (HR Friendly)
“A committee is just a group of people, so who is in it matters, but not what order they were picked in — that means nCr, not nPr. If someone must be on the committee, lock in that seat and choose the rest from everyone else. If the committee needs a fixed number from different subgroups, work out each subgroup’s combination separately and multiply the results together, since those choices happen independently.”
Follow-up Questions
- How would you solve this if two specific people cannot serve together?
- How do you handle a committee that needs at least 2 members from a subgroup?
- How does the approach change if the committee has distinct roles like chair and secretary?
- How would you count committees excluding at least one of two rival candidates?
MCQ Practice
1. A 5-person committee is chosen from 12 people. A specific person must be included. How many committees are possible?
Fix the required person; choose remaining 4 from 11: 11C4 = 330.
2. A committee of 4 is chosen from 6 men and 5 women, requiring exactly 2 men and 2 women. How many ways?
6C2 × 5C2 = 15 × 10 = 150.
3. A 3-person committee is chosen from 8 people, and a specific person must NOT be included. How many committees are possible?
Choose all 3 from the remaining 7 people: 7C3 = 35.
Flash Cards
Why does committee selection use nCr, not nPr? — Because membership is unordered — a group, not a sequence.
How to handle a required member? — Fix that member, then choose the rest from the reduced pool.
How to handle subgroup quotas (e.g., 2 men, 3 women)? — Compute each subgroup’s combination separately, then multiply.
How to handle “at least k from a subgroup”? — Sum valid cases, or subtract invalid cases from the unrestricted total.