Combinations: Explained with Examples, Concepts and Examples

A combination counts selections where the order does not matter: choosing 3 friends out of 10 for a team is a single combination no matter the order in which you name them. This is the mathematics behind lottery odds, poker hands and the classic "in how many ways" exam questions.

It matters because combinations separate which items are chosen from in what order — and confusing the two is the single most common counting error students make.

Key Concepts

• Combination C(n, r), written nCr: the number of ways to choose r items from n, order ignored.
• Permutation P(n, r): the number of ways to choose and arrange r items from n — order matters, and P(n, r) = C(n, r) × r!.
• Factorial: n! = n × (n−1) × … × 1, with 0! = 1 by definition.
• Symmetry: C(n, r) = C(n, n − r) — choosing r items to take is the same as choosing n − r to leave behind.
• Pascal's triangle: each entry is the sum of the two above it; the entries of row n are exactly C(n, 0), C(n, 1), …, C(n, n), and each row sums to 2ⁿ.

Formulas

• C(n, r) = n! ÷ (r! × (n − r)!)
• Useful values: C(n, 0) = C(n, n) = 1; C(n, 1) = n.

Worked Example

In how many ways can a committee of 3 be chosen from 10 students?

1. C(10, 3) = 10! ÷ (3! × 7!).
2. Cancel 7!: (10 × 9 × 8) ÷ (3 × 2 × 1) = 720 ÷ 6 = 120 ways.
3. Symmetry check: C(10, 7) — choosing 7 to leave out — must equal C(10, 3): 10! ÷ (7! × 3!) = 120 ✓.
4. Contrast with permutations: assigning the 3 chosen students as president, secretary and treasurer gives P(10, 3) = 10 × 9 × 8 = 720 = 120 × 6, the 6 possible orders of each chosen trio.

The 120 vs 720 comparison is the whole lesson: combinations count the teams, permutations count the teams plus their internal orderings.

Common Misconceptions

• "C(10,3) and P(10,3) are basically the same." Permutations also count orderings, so they are r! times larger — here, 6 times larger.
• "Order never matters in real problems." Dealing distinct roles or cards to named players is a permutation problem; forming an unnamed team is a combination problem. Read the question's wording.
• "C(n, r) always needs the full factorial." Use symmetry C(n, r) = C(n, n − r) to work with the smaller number, and cancel factorials before multiplying — C(50, 48) is just C(50, 2) = 1,225.

Real-World Applications

• Lottery odds: a 6-from-49 lottery has C(49, 6) = 13,983,816 possible tickets.
• Poker: C(52, 5) = 2,598,960 possible 5-card hands — the denominator of every poker probability.
• Quality-control sampling: choosing which items to inspect from a production batch.
• Binomial theorem: the coefficients of (a + b)ⁿ are exactly C(n, 0) … C(n, n).

A combination counts unordered selections with C(n, r) = n! ÷ (r!(n − r)!), symmetric as C(n, r) = C(n, n − r). Decide first whether order matters — teams and tickets are combinations, ranks and arrangements are permutations — and the counting becomes mechanical.