Combinations
Key Terms
- Combination C(n, r)
- The number of ways to choose r objects from n distinct objects where order does NOT matter; C(n, r) = n! / [r!(n−r)!].
- Order does not matter
- C(n, r) = P(n, r) / r! — dividing permutations by r! removes the r! orderings of each selection.
- Symmetry property
- C(n, r) = C(n, n−r) — choosing r objects to include is the same as choosing n−r to exclude.
- Special values
- C(n, 0) = C(n, n) = 1; C(n, 1) = n.
- Complementary counting
- Total − (unwanted outcomes) = wanted outcomes; often easier than counting directly.
- Pascal’s triangle
- Row n of Pascal’s triangle lists C(n,0), C(n,1), …, C(n,n) — each entry is the sum of the two above it.
The Combination Formula
C(n, r) counts the number of ways to choose r objects from n distinct objects where order does not matter:
C(n, r) = nCr = n! ÷ (r! × (n − r)!)
Key properties:
• C(n, 0) = C(n, n) = 1 • C(n, 1) = n • C(n, r) = C(n, n−r) (symmetry)
• Relationship to permutations: C(n, r) = P(n, r) ÷ r!
| Situation | Use | Reason |
|---|---|---|
| Choosing a committee of 3 from 10 | C(10, 3) = 120 | No roles; order irrelevant |
| Choosing president, secretary, treasurer from 10 | P(10, 3) = 720 | Distinct roles; order matters |
| Choose 2 boys AND 3 girls (from 5 boys, 4 girls) | C(5,2) × C(4,3) | Independent selections → multiply |
| Choose 3 from boys OR from girls | C(5,3) + C(4,3) | Mutually exclusive alternatives → add |
Worked Example 1 — Committee of 3 from 8
In how many ways can a committee of 3 be chosen from 8 people?
Order does not matter (a committee has no distinct roles) → use C(8, 3).
C(8, 3) = 8! ÷ (3! × 5!) = (8 × 7 × 6) ÷ (3 × 2 × 1) = 336 ÷ 6 = 56 committees
Worked Example 2 — 5 Cards from a Deck of 52
How many different 5-card hands can be dealt from a standard 52-card deck?
The order of cards in a hand does not matter → C(52, 5).
C(52, 5) = 52! ÷ (5! × 47!) = (52 × 51 × 50 × 49 × 48) ÷ (5 × 4 × 3 × 2 × 1) = 311 875 200 ÷ 120 = 2 598 960 hands
Worked Example 3 — Combined Selection
A team of 5 is chosen from 5 boys and 4 girls. How many teams contain exactly 2 boys and 3 girls?
Choose 2 from 5 boys: C(5, 2) = 10. Choose 3 from 4 girls: C(4, 3) = 4.
These are independent selections, so multiply:
Total = C(5, 2) × C(4, 3) = 10 × 4 = 40 teams
Why Divide by r! to Get Combinations?
Permutations count ordered arrangements: P(n, r) counts every way to pick r items from n and arrange them in a sequence. But for a combination, we only care which r items are chosen, not their order.
For any given selection of r items, there are r! different orderings — all of which count as the same combination. So we divide P(n, r) by r! to collapse all orderings of the same group into a single count:
C(n, r) = P(n, r) / r! = n! / (r! × (n−r)!)
Example: The committee {Alice, Bob, Carol} is the same regardless of whether we pick Alice first, Bob first, or Carol first. All 3! = 6 orderings represent the same committee, so we divide by 6.
The Symmetry Property: C(n, r) = C(n, n−r)
Choosing r items to include is equivalent to choosing n−r items to exclude. Every selection of r members from n corresponds uniquely to one selection of n−r members (the ones left out). This is the symmetry property.
Verification: C(8, 3) = 8!/(3!×5!) = 56. C(8, 5) = 8!/(5!×3!) = 56. ✓
AND vs OR in Multi-Selection Problems
AND (sequential independent choices): When you must select from two groups simultaneously (e.g., choose boys AND girls), treat each group separately and multiply the results. This is an application of the multiplication principle.
OR (mutually exclusive alternatives): When different configurations are possible (e.g., choose all from boys OR all from girls), count each case separately and add. The key requirement is that the cases are mutually exclusive.
Complementary counting: Sometimes it is easier to count the opposite and subtract from the total. If “at least one” or “at most” language appears, consider: Desired = Total − (Undesired).
Combinations with Restrictions
Must include: If a specific person must be on the team, include them first (1 way) and choose the remaining members from the rest.
Must exclude: Simply remove the excluded person from the pool before applying C(n, r).
Grouping restriction: If two people cannot both be on the committee, use: Valid = Total − (both included). Both included means fixing both and choosing the rest.
Distinguishing Combinations from Permutations
Ask: “Does swapping two selected items create a different outcome?”
• If yes (e.g., different roles, ordered list, sequence) → Permutation.
• If no (e.g., a group, committee, hand of cards, team) → Combination.
Mastery Practice
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Fluency
Q1 — Calculate Combinations
Evaluate: (a) C(7, 2) (b) C(10, 4) (c) C(9, 9) (d) C(12, 10)
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Fluency
Q2 — Choosing a Committee
In how many ways can a committee of 4 people be chosen from a group of 9 people?
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Fluency
Q3 — Selecting Players
A squad of 15 rugby players is available. In how many ways can 11 players be selected to start a match?
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Fluency
Q4 — Combinations vs Permutations
From a group of 10 students, decide whether to use C or P and calculate: (a) select 3 to represent the school at a debate (b) select a captain, vice-captain and manager.
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Understanding
Q5 — Mixed Gender Selection
A team of 5 is chosen from 6 boys and 5 girls. How many teams contain: (a) exactly 3 boys and 2 girls? (b) exactly 4 girls?
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Understanding
Q6 — Card Hands
Five cards are drawn from a standard 52-card deck (13 cards in each of 4 suits). How many 5-card hands contain: (a) exactly 3 aces? (b) all 5 cards from the same suit (a flush)?
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Understanding
Q7 — Committees with a Required Member
A committee of 5 is formed from 12 people. Person A must be on the committee. In how many ways can the committee be formed?
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Understanding
Q8 — Committees with an Excluded Pair
A committee of 4 is to be chosen from 10 people. Two people, Tom and Jerry, refuse to serve together. How many valid committees exist?
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Problem Solving
Q9 — Selecting at Least One
From a group of 4 teachers and 6 students, a committee of 4 is formed. How many committees contain at least one teacher?
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Problem Solving
Q10 — Handshakes in a Group
At a meeting, every person shakes hands exactly once with every other person. If there are 45 handshakes in total, how many people are at the meeting?