Combinations & Permutations Calculator
nCr and nPr — counting selections with and without order.
Permutations ₙPᵣ = 720 (order matters)
Use a combination when order doesn’t matter and a permutation when it does. To choose 3 from 10, ₁₀C₃ = 10! ÷ (3! · 7!) = 120; if order matters, ₁₀P₃ = 10! ÷ 7! = 720 — exactly 3! = 6 times larger.
Does order matter?
That single question decides which formula to use. A combination (ₙCᵣ) counts selections where order is irrelevant — choosing 3 toppings from 10. A permutation (ₙPᵣ) counts arrangements where order matters — the top 3 finishers in a race. Both are built from factorials.
combinations ignore order; permutations count it, so ₙPᵣ = ₙCᵣ × r!
Worked example
Choose 3 from 10, with order ignored — ₙCᵣ where n = 10, r = 3:
- 1 Write the formula. ₁₀C₃ = 10! ÷ (3! · 7!). Cancel the 7! to get (10 × 9 × 8) ÷ 3!.
- 2 Compute top and bottom. Top: 10 × 9 × 8 = 720. Bottom: 3! = 6.
- 3 Divide. 720 ÷ 6 = 120. If order mattered, ₁₀P₃ = 720 — exactly 3! = 6 times larger.
Small combinations (ₙCᵣ)
Pascal’s triangle: each value is the number of ways to choose r items from n. Rows are symmetric.
| n \ r | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| 2 | 1 | 2 | 1 | — | — | — |
| 3 | 1 | 3 | 3 | 1 | — | — |
| 4 | 1 | 4 | 6 | 4 | 1 | — |
| 5 | 1 | 5 | 10 | 10 | 5 | 1 |
The link, and common cases
Permutations = combinations × r!. Every combination can be ordered in r! ways, so ₙPᵣ = ₙCᵣ × r!. Permutations are always at least as many as combinations.
ₙCᵣ = 1 at the edges. Choosing all n items, or choosing none, can be done exactly one way — which is why ₙC₀ = ₙCₙ = 1.
Symmetry. ₙCᵣ = ₙC₍ₙ₋ᵣ₎: choosing which 3 to take is the same as choosing which 7 to leave, so ₁₀C₃ = ₁₀C₇ = 120.