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Permutation and Combination Calculator

By Calxy · Updated · Calculation methodology

Count the ways to arrange items when order matters (permutations) or to choose a group when it does not (combinations), with and without repeats.

Printed from Calxy · https://www.calxy.net/math/permutation-combination-calculator

How many things you can choose from, such as 10 people.

Combinations C(10, 3)

120

Permutations P(10, 3) = 720

  • Permutations, no repetition — nPr = n! ÷ (n − r)!720
  • Combinations, no repetition — nCr = n! ÷ (r!(n − r)!)120
  • Permutations with repetition — n^r1,000
  • Combinations with repetition — C(n + r − 1, r)220
  • r! (arrangements of each chosen set)6
  • Circular permutations — P(n, r) ÷ r240
  • Even / odd permutations of n — n! ÷ 2 each1,814,400 / 1,814,400
All r for n = 10
rnPrnCr
011
11010
29045
3720120
45,040210
530,240252
6151,200210
7604,800120
81,814,40045
93,628,80010
103,628,8001

How this was calculated

  1. A permutation is an arrangement where order matters (ABC is different from CBA). There are 10 choices for the first place, 9 for the second, and so on for 3 places. Multiply them. (The ! sign means “multiply down to 1”, so 4! = 4 × 3 × 2 × 1.)P(n, r) = n! ÷ (n − r)!Put in your numbers: P(10, 3) = 10! ÷ 7! = 10 × 9 × 8Answer: 720
  2. A combination is a group where order does not matter (ABC is the same as CBA). Each group of 3 can be put in order 3! ways, so divide the permutations by 3!.C(n, r) = P(n, r) ÷ r!Put in your numbers: C(10, 3) = 720 ÷ 3! = 720 ÷ 6Answer: 120Each unordered selection can be ordered 3! ways.
  3. If an item can be picked again (repetition), each of the 3 places has all 10 choices, so multiply 10 by itself 3 times. A 4-digit PIN works this way: 10^4 = 10,000.With repetition, ordered: n^rPut in your numbers: 10^3 = 10 × 10 × 10Answer: 1,000
  4. Picking with repetition when order does not matter, like choosing 3 scoops from 10 ice-cream flavours, is the same as choosing 3 from 12 (the “stars and bars” trick).With repetition, unordered: C(n + r − 1, r)Put in your numbers: C(10 + 3 − 1, 3) = C(12, 3)Answer: 220
  5. Around a circle, moving everyone one seat along gives the same arrangement. Each circle of 3 was counted 3 times in the line count, so divide by 3.Circular arrangements of r from n: P(n, r) ÷ rPut in your numbers: 720 ÷ 3Answer: 240Rotations of the same circle count once. If flipping the circle over also counts as the same (a necklace), halve this for r ≥ 3.
  6. Every ordering of all n items can be made from the starting order by swapping two items at a time. It is even if that takes an even number of swaps and odd if it takes an odd number. Exactly half are each.Even and odd permutations of n elements: n! ÷ 2 each (n ≥ 2)Put in your numbers: 10! ÷ 2 = 3,628,800 ÷ 2Answer: 1,814,400 even, 1,814,400 odd

Permutations and combinations in simple words

Both count the ways to pick some things from a group. The difference is whether order matters.

A permutation is when order matters, like first, second and third place in a race. With 10 runners, there are 10 choices for first, then 9 for second, then 8 for third: 10 × 9 × 8 = 720 ways.

A combination is when order does not matter, like picking 3 friends to come to the park. Picking Ann, Ben and Cat is the same group as Cat, Ben and Ann. Each group of 3 can be put in order 3 × 2 × 1 = 6 ways, so there are 720 ÷ 6 = 120 groups.

So there are always at least as many permutations as combinations, because every group can be lined up in several orders.

Does order matter?

Permutations count ordered arrangements: choosing A then B differs from B then A. Combinations count unordered selections, so those choices are the same. Without repetition, P(n, r) = n! ÷ (n − r)! and C(n, r) = n! ÷ [r!(n − r)!].

With repetition allowed, ordered choices give n^r. Unordered choices give C(n + r − 1, r) for n > 0. These formulas assume distinct item types and no restrictions beyond the chosen repetition rule.

Example: choose 3 from 10

Without repetition, there are 10 × 9 × 8 = 720 ordered arrangements. Each group of three has 3! = 6 orders, so there are 720 ÷ 6 = 120 combinations.

With repetition, ordered choices give 10³ = 1,000. Unordered choices give C(12, 3) = 220.

Empty selections and large counts

Choosing zero items has one outcome: the empty selection. Choosing more items than are available has zero outcomes without repetition. From zero item types, choosing a positive number of items is impossible even with repetition.

Enter whole numbers n and r between 0 and 1,000. Arithmetic uses big integers; results longer than 24 digits are displayed in abbreviated scientific notation. That display is approximate even though the underlying count is exact.

Circular, even and odd permutations

Around a circle, rotations of the same arrangement count once. Seating all n people gives (n − 1)! arrangements, so 5 people can sit 4! = 24 ways. Seating r of n people gives P(n, r) ÷ r: choosing 3 of 5 gives 60 ÷ 3 = 20.

A permutation is even or odd according to whether it takes an even or odd number of swaps to produce. For n ≥ 2, exactly half of the n! permutations are even: 4 elements have 12 even and 12 odd permutations.

Frequently asked questions

When should I use permutations?

Use permutations when position or sequence changes the outcome, such as assigning first, second and third prizes. Use combinations for a committee where membership matters but order does not.

What does repetition mean?

The same item type may be chosen more than once. A three-digit code using ten possible digits has 10³ ordered outcomes when repeated digits are allowed.

Is a necklace the same as a circular permutation?

Not quite. A necklace can also be flipped over, so each arrangement matches its mirror image. For three or more beads, divide the circular count by 2.

References

Last updated . Results are estimates for informational purposes only.