Permutation and Combination Calculator

Permutations, nPr
Order matters
720
10P3 = 10 × 9 × 8 = 720
Combinations, nCr
Order doesn't matter
120
10C3 = 10P3 ÷ 3! = 720 ÷ 6 = 120

With repetition allowed

Arrangements, nʳ
Order matters
1,000
10³
Selections, (n + r − 1)Cr
Order doesn't matter
220
12C3

What are permutations and combinations?

Both count the ways to choose r items from n. In a permutation the order matters, as with first, second and third prizes or a PIN. In a combination it doesn't, as with picking a team or a set of questions to answer. Because each group of r can be arranged in r! orders, there are always r! times as many permutations as combinations.

nPr and nCr formulas

nPr = n! ÷ (n − r)! · nCr = n! ÷ (r! × (n − r)!) = nPr ÷ r! · With repetition: nʳ arrangements, (n + r − 1)Cr selections

n! (n factorial) is n × (n − 1) × … × 1, and 0! = 1. With repetition allowed, the same item can be chosen more than once, like the digits of a PIN.

Permutation or combination?

  • Ask whether swapping two chosen items gives a different result. If it does (ranks, seats, passwords), use permutations; if not (teams, hands of cards), use combinations.
  • nCr = nC(n − r): choosing 3 to include from 10 is the same as choosing 7 to leave out, so 10C3 = 10C7 = 120.
  • nC0 = nCn = 1 and nC1 = n.
  • Arranging all n items in a row gives n! ways: 5 books can be shelved in 5! = 120 orders.
  • A 5-card hand from a 52-card deck can be any of 52C5 = 25,98,960 combinations.

Worked example: choosing 3 people from 10

  1. A club has 10 members and needs a president, a secretary and a treasurer
  2. Order matters (the roles differ): 10P3 = 10 × 9 × 8 = 720 ways
  3. For a committee of 3 with no roles, order doesn't matter
  4. 10C3 = 10P3 ÷ 3! = 720 ÷ 6 = 120 ways
  5. A 3-digit code from the digits 0–9, repeats allowed: 10³ = 1,000 codes

Common nPr and nCr values

Values that come up in textbook and exam questions.

nrnPrnCr
42126
536010
623015
8333656
103720120
11555,440462
12411,880495
52531,18,75,20025,98,960

nPr is always nCr × r!.

Frequently Asked Questions

A permutation counts ordered arrangements and a combination counts unordered selections. Choosing 3 of 10 people gives 720 permutations (for 3 different roles) but 120 combinations (for a committee).
Divide n! by r! × (n − r)!. A quicker way: multiply r numbers counting down from n, then divide by r!. 10C3 = (10 × 9 × 8) ÷ (3 × 2 × 1) = 120.
0! = 1 by definition. This makes formulas such as nCn = n! ÷ (n! × 0!) = 1 work: there's exactly one way to choose everything.
Not without repetition: you can't choose 5 different items from 3. With repetition it can, since items may be reused; 3 digits can make 3⁵ = 243 five-digit codes.