Permutation and Combination Calculator

Math

Calculate permutations nPr and combinations nCr with or without repetition. Features exact BigInt integer calculation for large numbers and step-by-step factorial reductions.

Presets:

Permutations: nPr

Order Matters (Sequences)

720

P(10, 3)

Combinations: nCr

Order Does NOT Matter (Groups)

120

C(10, 3)

🧮 Combinatorics Matrix Summary (n = 10, r = 3)

Condition	Permutations (Order Matters)	Combinations (Order Does NOT Matter)
Without Repetition	720	120
With Repetition Allowed	1,000	220

Step-by-Step Mathematical Derivations

1 1. Permutations: Order Matters

nPr = n! / (n - r)! = 10! / (10 - 3)! = 720

The number of ordered arrangements of 3 items chosen from 10 distinct items.

2 2. Combinations: Order Does NOT Matter

nCr = n! / [r! · (n - r)!] = C(10, 3) = 120

The number of subsets of size 3 selected from 10 items without regard to order.

3 3. Permutation vs Combination Relationship

nPr = nCr × r! = 120 × 3! = 720

For any combination of r items, there are r! ways to arrange them in order.

How This Calculator Works

Permutations and combinations are core counting techniques in combinatorics. A permutation calculates the number of ways to arrange objects where order matters (such as digits in a passcode). A combination calculates the number of ways to select subsets of objects where order does not matter (such as lottery numbers).

📐 Combinatorics Formulas Matrix

nPr = n! / (n - r)! | nCr = n! / [r! · (n - r)!] | nPr(rep) = n^r | nCr(rep) = (n + r - 1)! / [r!(n - 1)!]

n: Total number of items in the population set

r: Number of items selected or arranged

n!: Factorial: n × (n - 1) × ... × 2 × 1 (with 0! = 1)

nPr: Permutations without replacement

nCr: Combinations without replacement

Key Mathematical Rules & Properties

Does Order Matter?

If switching the order of chosen elements creates a new distinct outcome, use Permutations (nPr). If {A, B} is identical to {B, A}, use Combinations (nCr).

Combinations Symmetry Property

Choosing r items out of n leaves exactly n - r items behind, so C(n, r) is always equal to C(n, n - r). For example, C(10, 8) = C(10, 2) = 45.

✎

Worked Example: Selecting Committee Members

How many ways can a 3-person subcommittee be selected from a board of 8 members?

1. Order does not matter for committee membership → use Combinations (nCr).
2. n = 8, r = 3.
3. nCr = 8! / [3! · (8 - 3)!] = 8! / (3! · 5!).
4. Cancel 5!: (8 × 7 × 6) / (3 × 2 × 1) = 336 / 6 = 56 ways.

Final Answer: 56 possible subcommittees

Frequently Asked Questions

Why is 0! equal to 1?

By convention and algebraic consistency, there is exactly one way to arrange zero objects: the empty set {}. Also, setting 0! = 1 ensures that n! / (n - 0)! = 1 correctly.

Understanding your calculation

Method, example and limitations

How the calculation works

Permutations count ordered selections, n!/(n − r)!; combinations count unordered selections, n!/[r!(n − r)!], when repetition is excluded.

Example to check

Choosing two people from five gives 10 combinations. Assigning two distinct roles from five people gives 20 permutations.

Before using the result

Check whether order matters and whether repetition is allowed. The formulas require nonnegative integers with r ≤ n.

Calculation methodology · Report a result that needs review