CALTRX

Permutation and Combination Calculator

Enter n and r to compute both permutations and combinations. Get accurate results with step-by-step explanations.

n

Enter your values above to see the results.

Tips & Notes

  • ✓Key question: does order matter? YES → permutation. NO → combination.
  • ✓C(n,r) = C(n, n−r). Choosing 3 from 10 = leaving 7 from 10. Use smaller r to compute.
  • ✓P(n,r) counts down r terms from n: P(8,3) = 8×7×6. Stop after r terms.
  • ✓For combinations with repetition: C(n+r−1, r). Choosing 3 flavors from 5 with repeats: C(7,3)=35.
  • ✓Multiply probabilities: P(picking specific 3 from 8 in order) = 1/P(8,3) = 1/336.

Common Mistakes

  • ✗Using permutation formula when order does not matter. Choosing 3 committee members from 10 people: use C(10,3) = 120, not P(10,3) = 720. P counts each arrangement of the same 3 people as different.
  • ✗Forgetting to divide by r! when converting permutation to combination. C(n,r) = P(n,r) / r!. For C(5,2): P(5,2) = 20, divide by 2! = 2 → C(5,2) = 10. Omitting the division gives 20, which overcounts.
  • ✗Confusing n and r in the formula. In C(n,r), n is the total pool and r is the number chosen. C(3,10) is undefined (cannot choose 10 from 3). Always verify n ≥ r before computing.
  • ✗Applying combination when repetition is allowed. Standard C(n,r) assumes no repetition. If the same item can be chosen multiple times, use the repetition formula C(n+r−1, r) instead.
  • ✗Computing factorials incorrectly for large numbers. 10! = 3,628,800 — not 10 × 9 = 90. For large n, use the multiplicative form: P(10,3) = 10 × 9 × 8 = 720 directly rather than computing full factorials.

Permutation and Combination Calculator Overview

Permutations and combinations answer the foundational question in combinatorics: how many ways can items be selected and arranged from a set? The critical distinction between the two is whether order matters. When different arrangements of the same items count as different outcomes, you use permutations. When only which items were selected matters — and arrangement is irrelevant — you use combinations.

Permutations count ordered selections of r items chosen from n distinct items:

P(n, r) = n! / (n − r)!
EX: Arrange 3 books chosen from a shelf of 8: P(8, 3) = 8 × 7 × 6 = 336 ordered arrangements
You multiply exactly r terms descending from n — no need to compute full factorials. Eight choices for the first position, seven remaining for the second, six for the third. Combinations count unordered selections — the same group regardless of the order it was assembled:
C(n, r) = n! / (r! × (n − r)!)
EX: Choose 3 people from a group of 8 for a committee: C(8, 3) = 56 possible groups
Combinations divide permutations by r! to eliminate the arrangements that permutations treat as distinct. The committee {Alice, Bob, Carol} is the same group regardless of the order they were selected. These two quantities are always linked by a precise relationship:
P(n, r) = C(n, r) × r!
EX: P(8, 3) = C(8, 3) × 3! = 56 × 6 = 336 ✓
Permutations count arrangements when order matters; combinations count selections when order does not. The key question before calculating: does rearranging the elements produce a different outcome? Arranging books on a shelf — yes, order matters. Choosing committee members — no, the same people in a different order form the same committee. Getting this distinction right determines whether P(n,r) or C(n,r) applies. Several identities simplify large computations. C(n, k) = C(n, n−k): choosing k items is equivalent to choosing which n−k items to leave out. C(n, 0) = C(n, n) = 1: there is exactly one way to choose nothing and exactly one way to choose everything. P(n, n) = n!: arranging every item in a set produces n! distinct orderings. These identities reduce calculation effort and reveal the symmetry in counting problems that would otherwise require full factorial arithmetic.

Frequently Asked Questions

A permutation counts ordered arrangements — the sequence matters. A combination counts unordered selections — only which items are chosen matters. Example: from {A, B, C}, selecting 2. Permutations: AB, BA, AC, CA, BC, CB — 6 arrangements. Combinations: AB, AC, BC — 3 groups. The same 3 people arranged differently still give the same committee (combination), but a different batting order (permutation). Question to ask: would swapping two chosen items give a different outcome? If yes, use permutation; if no, use combination.

P(n,r) = n!/(n−r)! counts ordered arrangements of r items from n. C(n,r) = n!/(r!(n−r)!) counts unordered selections. Example: arrange 3 of 5 books on a shelf: P(5,3) = 5×4×3 = 60 arrangements. Choose 3 of 5 books (order irrelevant): C(5,3) = 10 selections. C(5,3) = P(5,3)/3! = 60/6 = 10 — dividing by r! removes the overcounting of different orderings of the same group. This ratio is why combinations are always smaller than or equal to the corresponding permutation.

Combinations follow Pascal's identity: C(n,r) = C(n−1,r−1) + C(n−1,r). Each entry in Pascal's triangle equals the sum of the two entries above it. Row 5: 1, 5, 10, 10, 5, 1 → C(5,0) through C(5,5). The binomial theorem states (a+b)ⁿ = Σ C(n,k)aⁿ⁻ᵏbᵏ. Example: (x+y)³ = C(3,0)x³ + C(3,1)x²y + C(3,2)xy² + C(3,3)y³ = x³+3x²y+3xy²+y³. Combinations are the binomial coefficients — they count outcomes and expand polynomials.

Permutations with repetition: when items can be reused, the count is nʳ. Example: 4-digit PIN using digits 0–9 with repetition: 10⁴ = 10,000 possibilities. Combinations with repetition (choosing r from n with replacement, order irrelevant): C(n+r−1, r). Example: choosing 3 ice cream scoops from 5 flavors with repetition allowed: C(5+3−1, 3) = C(7,3) = 35 combinations. Without repetition: C(5,3) = 10. Allowing repetition always gives more possibilities than without.

Probability uses counting: P(event) = favorable outcomes / total outcomes. Example: probability of getting exactly 2 heads in 5 coin flips. Total outcomes = 2⁵ = 32. Favorable: C(5,2) = 10 ways to choose which 2 flips are heads. P(exactly 2 heads) = 10/32 = 5/16 ≈ 31.25%. This uses the binomial probability formula: P(X=k) = C(n,k)×pᵏ×(1−p)ⁿ⁻ᵏ = C(5,2)×0.5²×0.5³ = 10×0.03125 = 0.3125.

Circular permutations fix one element to remove rotational equivalence. n distinct objects around a circle: (n−1)! arrangements (not n!). Example: seating 5 people at a round table: (5−1)! = 24 arrangements (not 5!=120, because rotations of the same arrangement are equivalent). If the circle also has reflections considered equivalent (like a necklace): (n−1)!/2. 5-bead necklace: (5−1)!/2 = 12 distinct patterns. This reduction factor of n (for circles) or 2n (for necklaces) accounts for symmetry.