This free online permutation and combination calculator finds both nPr (the number of ways to arrange r items from a set of n in order) and nCr (the number of ways to choose r items from n when order doesn't matter). It's widely used in probability, statistics, and combinatorics problems.
How to Use the Permutation and Combination Calculator
Enter the total number of items in the "Total number of items (n)" field.
Enter how many items you're choosing in the "Number of items chosen (r)" field.
Click "Calculate" to see both the nPr (permutations) and nCr (combinations) results.
The Formulas and How They Connect
nPr = n! ÷ (n − r)!, the number of ordered arrangements of r items taken from n.
nCr = n! ÷ (r! × (n − r)!), the number of unordered selections.
The two are linked by nCr = nPr ÷ r!. Every group of r chosen items can be lined up in r! different orders, so the permutation count is always r! times the combination count. With n = 10 and r = 3, nPr = 10 × 9 × 8 = 720 and nCr = 720 ÷ 6 = 120.
Worked Examples: Does Order Matter?
Situation
n, r
Use
Answer
Elect a class president, vice-president and secretary from 10 students
10, 3
nPr
720
Pick a 3-member committee from the same 10 students
10, 3
nCr
120
Choose a playing XI from a cricket squad of 15
15, 11
nCr
1,365
Form a 4-digit code using different digits 0–9
10, 4
nPr
5,040
Number of possible 5-card hands from a 52-card deck
52, 5
nCr
2,598,960
A quick test: if swapping two chosen items gives a different outcome (different posts, different digit positions, different finishing places), order matters and you need nPr. If the result is just a group, use nCr.
What This Calculator Does Not Cover
Repetition allowed. If a digit or item can be reused, as in an ordinary 4-digit PIN, the count is n to the power r (10,000 PINs), not nPr.
Identical items. Arranging the letters of a word with repeated letters, such as BANANA, needs n! divided by the factorial of each repeat count: 6! ÷ (3! × 2!) = 60.
Circular arrangements. Seating n people around a round table gives (n − 1)! arrangements, because rotations count as the same.
Very large answers. Results are computed with ordinary floating-point numbers, which store about 15 to 16 significant digits. Beyond about 9 × 10¹⁵ the trailing digits are rounded; for example, 100C50 is shown as 100,891,344,545,564,200,000,000,000,000 while the exact value ends in ...193,334,812,497,256. Results that overflow completely show a "too large" message.
Frequently Asked Questions
A permutation (nPr) counts arrangements where order matters, while a combination (nCr) counts selections where order doesn't matter. For example, choosing 2 letters from the set A, B, C gives nPr = 6 ordered pairs but only nCr = 3 unordered pairs.
It computes nPr = n × (n − 1) × ... × (n − r + 1), then derives nCr = nPr / r! by dividing the permutation count by r factorial. Both are computed iteratively rather than with full factorials, so larger values of n won't overflow as quickly.
There is exactly one way to choose nothing, so nC0 = 1, and the empty arrangement also makes nP0 = 1. Choosing r items is the same as choosing which n − r items to leave out, so nCr always equals nC(n − r); both 5C2 and 5C3 are 10.
Divide the number of favourable selections by the total number of selections. For example, there are 49C6 = 13,983,816 ways to pick 6 numbers from 49, so the chance that one ticket matches all six drawn numbers is 1 in 13,983,816.
Without repetition you cannot pick more items than there are, so there are zero ways to do it. If items can be reused, r may exceed n, but that case needs the formula n to the power r instead of nPr.