Combination and Permutation Calculator
Calculate combinations (nCr) and permutations (nPr) with or without repetition. Shows all four counts at once, factorials, probability, and a row of Pascal's triangle.
Result
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Formula —
Extended More scenarios, charts & detailed breakdown ▾
nCr (Combinations)
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Formula —
Professional Full parameters & maximum detail ▾
Core Counts
Combinations nCr —
Permutations nPr —
Combinations with Repetition —
Permutations with Repetition (nʳ) —
Factorials
n! (Factorial of n) —
r! (Factorial of r) —
Probability
Probability (nCr / total outcomes) —
Pascal's Triangle
Pascal's Triangle Row n (first 8 shown) —
How to Use This Calculator
Enter n (total items) and r (items chosen) and select whether you want Combinations or Permutations. The result and formula display instantly. Use the Extended tabs to toggle between nCr, nPr, and versions with repetition. The Professional tab shows all four counts, factorials, probability, and a row of Pascal's triangle.
Formula
nCr = n! / (r! × (n−r)!) | nPr = n! / (n−r)!
With repetition (combos): (n+r−1)! / (r!(n−1)!) | With repetition (perms): nʳ
Example
n=10, r=3: nCr = 120, nPr = 720, Combos w/ rep = 220, Perms w/ rep = 1000
Frequently Asked Questions
- The key distinction is whether order matters. A combination counts selections where order does not matter: choosing 3 people from 10 for a committee gives the same committee regardless of what order you pick them. A permutation counts arrangements where order matters: awarding 1st, 2nd, and 3rd place from 10 contestants gives different results depending on order. So P(10,3) = 720 distinct ordered arrangements, but C(10,3) = 120 unordered groups — the ratio P/C = 3! = 6 because each group of 3 can be arranged 6 different ways.
- C(n, r) = n! / (r! × (n−r)!), where n! = n × (n−1) × … × 2 × 1. This counts the number of ways to choose r items from n without regard to order. For C(10, 3): 10! / (3! × 7!) = (10 × 9 × 8) / (3 × 2 × 1) = 720 / 6 = 120. A useful shortcut: compute only the top r terms of n! in the numerator and divide by r!. For large n, use the multiplicative formula: C(n,r) = (n/1) × ((n−1)/2) × ((n−2)/3) × … × ((n−r+1)/r).
- P(n, r) = n! / (n−r)!, counting the number of ways to arrange r items chosen from n where order matters. For P(10, 3): 10! / 7! = 10 × 9 × 8 = 720. The difference from combinations is that each unordered selection of r items contributes r! ordered arrangements, so P(n,r) = C(n,r) × r!. Example: P(5,2) = 5×4 = 20 ordered pairs; C(5,2) = 10 unordered pairs. Permutations are used in lock codes, race rankings, and seating arrangements.
- When items can be selected more than once (with replacement), the count changes. The number of ways to choose r items from n types with repetition allowed is C(n+r−1, r) = (n+r−1)! / (r! × (n−1)!). Example: choosing 3 scoops from 5 ice cream flavors where you can repeat flavors: C(5+3−1, 3) = C(7, 3) = 35. Without repetition (regular combination): C(5,3) = 10. The "stars and bars" technique provides the intuition for the with-repetition formula.
- Pascal's triangle is a number triangle where each entry is the sum of the two entries directly above it. The nth row (starting from row 0) contains the binomial coefficients C(n,0), C(n,1), …, C(n,n), which match the combination values. Row 0: [1]. Row 1: [1, 1]. Row 2: [1, 2, 1]. Row 3: [1, 3, 3, 1]. Row 4: [1, 4, 6, 4, 1]. The entries also equal the coefficients in the binomial expansion (a+b)^n. Pascal's triangle is related to the Fibonacci sequence, the Sierpinski triangle fractal, and many combinatorial identities.