Permutation and Combination Calculator
Compute nPr, nCr, n!, arrangements with repetition and multiset permutations exactly, with the formula substituted step by step and checked in a second way.
Permutations versus combinations
Counting problems come down to two questions: does the order of selection matter, and may items be repeated? A permutation counts ordered arrangements, a combination counts unordered selections. Choosing a president, a secretary and a treasurer from ten people is a permutation problem, P(10, 3) = 720, because roles differ. Choosing a committee of three from ten people is a combination problem, C(10, 3) = 120, because only the group matters. Notice that 720 = 120 × 3!: each committee can be ordered in 3! ways.
The formulas
The calculator supports six counting situations.
- Factorial n! = n(n − 1)…1, the number of ways to arrange n distinct objects in a row; 0! = 1.
- Permutation P(n, r) = n! / (n − r)! for ordered selections without repetition.
- Combination C(n, r) = n! / (r!(n − r)!), also called a binomial coefficient, for unordered selections without repetition.
- Permutation with repetition nr, for example four-digit PIN codes: 104 = 10 000.
- Combination with repetition C(n + r − 1, r), from the stars-and-bars argument, for example choosing 3 scoops from 5 flavours with repeats allowed.
- Multiset permutation n! / (k₁! k₂! …), arrangements of items where some are identical, such as the letters of MISSISSIPPI.
Exact results for huge numbers
Counts grow explosively: 52! has 68 digits and 100! has 158. An ordinary calculator switches to scientific notation and loses digits, but this tool uses arbitrary-precision integers, so every digit is exact. To keep results readable, n is limited to 1000. For binomial coefficients the calculator never computes the full factorials; it cancels common factors, so even C(1000, 500) is instant.
Built-in cross-checks
Every answer is verified a second way: P(n, r) is compared with C(n, r) × r!, combinations are checked against symmetry C(n, r) = C(n, n − r) and Pascal's rule C(n, r) = C(n − 1, r − 1) + C(n − 1, r), and multiset counts against a group-by-group product of binomial coefficients. A mismatch would stop the calculation instead of showing a wrong number.
Worked examples
These examples are computed by the same engine as the calculator above, so you can compare each step with your own working.
C(10, 3)
- Counting type:
Combination C(n, r) - n:
10 - r:
3 - Group sizes (identical-items type):
1
Step-by-step solution
Formula used
C(n, r) = n! / (r!(n − r)!)
Order does not matter, no repetition. C(n, r) = C(n, n − r).
CalculateApply the combination formula
C(n, r) = n! / (r!(n − r)!)
C(10, 3) = 10! / (3! × 7!)
CalculateCancel the common factorial
C(10, 3) = (10 × 9 × 8) / 3!
= 720 / 6 = 120
ResultResult
C(10, 3) = 120
CheckCross-check
Symmetry: C(n, n − r) = 120 ✓
Pascal's rule: C(n − 1, r − 1) + C(n − 1, r) = 36 + 84 = 120 ✓
P(10, 3)
- Counting type:
Permutation P(n, r) - n:
10 - r:
3 - Group sizes (identical-items type):
1
Step-by-step solution
Formula used
P(n, r) = n! / (n − r)! = n(n − 1)…(n − r + 1)
Order matters, no repetition.
CalculateApply the permutation formula
P(n, r) = n! / (n − r)!
P(10, 3) = 10! / 7!
= 10 × 9 × 8
ResultResult
P(10, 3) = 720
CheckCross-check
n!/(n−r)! gives the same number ✓
C(n, r) × r! = 120 × 6 = 720 ✓ (choose first, then arrange)
52!
- Counting type:
Factorial n! - n:
52 - r:
0 - Group sizes (identical-items type):
1
52!
80658175170943878571660636856403766975289505440883277824000000000000
68 digits
Step-by-step solution
Formula used
n! = n × (n − 1) × … × 2 × 1, 0! = 1
CalculateMultiply the integers from 1 to n
n! = n·(n−1)·…·1
52! = 52 × 51 × 50 × 49 × … × 2 × 1
ResultResult
52! = 80658175170943878571660636856403766975289505440883277824000000000000
CheckCross-check
52! = 52 × 51! = 52 × 1551118753287382280224243016469303211063259720016986112000000000000 (a 67-digit number) ✓
MISSISSIPPI letters
- Counting type:
Arrangements with identical items - n:
0 - r:
0 - Group sizes (identical-items type):
1, 4, 4, 2
11! / (1! × 4! × 4! × 2!)
34650
Example: the word MISSISSIPPI has groups M=1, I=4, S=4, P=2, so its letters can be arranged in 1, 4, 4, 2 → 34650 ways.
Step-by-step solution
Formula used
Arrangements of n items with repeated groups k₁, …, kₘ: n! / (k₁! k₂! … kₘ!)
CalculateDivide by the arrangements inside each group
n! / (k₁! k₂! … kₘ!)
n = 1 + 4 + 4 + 2 = 11
11! / (1! × 4! × 4! × 2!) = 34650
ResultResult
11! / (1! × 4! × 4! × 2!) = 34650
CheckCross-check
Choosing positions group by group: C(11, 1) × C(10, 4) × C(6, 4) × C(2, 2) = 34650 ✓
3 scoops from 5 flavours
- Counting type:
Combination with repetition - n:
5 - r:
3 - Group sizes (identical-items type):
1
Step-by-step solution
Formula used
Multisets of size r from n types (stars and bars): C(n + r − 1, r)
CalculateStars and bars
C(n + r − 1, r)
Place 3 identical items into 5 types using 4 dividers.
C(5 + 3 − 1, 3) = C(7, 3) = 35
ResultResult
C(5 + 3 − 1, 3) = 35
CheckCross-check
Equivalently the number of non-negative integer solutions of x₁ + … + x5 = 3.
Frequently asked questions
What is the difference between nPr and nCr?
nPr counts ordered arrangements and nCr counts unordered selections. They are related by P(n, r) = C(n, r) × r!.
How many ways can 5 people sit in a row?
5! = 120. The first seat has 5 choices, the next 4, and so on.
What is 0!?
By definition 0! = 1: there is exactly one way to arrange no objects. This also makes the formulas for C(n, 0) and C(n, n) equal to 1.
When should I use stars and bars?
When you distribute identical items into distinct groups, or choose with repetition and without regard to order. The count is C(n + r − 1, r).