Permutation and Combination Calculator

Count ordered arrangements and unordered selections, with the order decision made before the factorials.

Drag & drop or upload an image or PDF

Enter n and r, such as compare 8P3 and 8C3

Enter nonnegative integers n and r with r no greater than n. State whether order matters, and mention repetition or identical items.

Try an example

Order is the deciding question

A permutation counts arrangements, so choosing Ana then Ben differs from choosing Ben then Ana. A combination counts groups, so those two orders represent the same selection.

The familiar nPr and nCr formulas assume distinct items selected without replacement. If repetition is allowed or some items are identical, say so in the problem because the counting model changes.

How to use the permutation and combination calculator

Enter the problem as written

Type or paste the full expression. You can also upload a clear photo or PDF and check the extracted text before solving.

Read the working, not only the answer

Each transformation is separated and explained so you can compare it with your own method.

Ask about any step

Continue in the same solution to request another method, check a restriction, or ask why a rule applies.

Permutation and combination formulas

Permutations keep the order of the selected items. Combinations divide by r factorial because the r selected items can be reordered without creating a new group.

_{n} P_{r} = \frac{n !}{( n - r )!},_{n} C_{r} = (r n) = \frac{n !}{r ! ( n - r )!}

A reliable way to work through it

Ask whether order matters

Ranks, roles, codes, and seatings usually use permutations. Committees and unordered groups use combinations.

Check replacement and repetition

The basic nPr and nCr formulas select distinct items without replacement. Repeated choices require a different rule.

Cancel factorials efficiently

Expand only the factors that survive cancellation, then return the exact whole-number count.

Worked example

From 8 students, count an ordered slate of 3 officers and an unordered 3-person committee

_{8} P_{3} = \frac{8 !}{5 !} = 8 \cdot 7 \cdot 6 = 336

The three officer roles are different, so order matters.

_{8} C_{3} = \frac{8 !}{3 ! 5 !} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56

A committee has no role order, so divide out the 3 factorial arrangements.

_{8} P_{3} = 3!_{8} C_{3}

Each committee can be assigned to the three officer roles in 3 factorial ways.

There are 336 ordered officer slates and 56 unordered 3-person committees.

Common mistakes to check

Ignoring order

A choice of people and an assignment to distinct roles are different counting problems.

Using r greater than n

Without replacement, the number selected cannot exceed the number of available distinct items.

Forgetting repetition changes the rule

Codes with repeated symbols and arrangements with identical items are not handled by the basic nPr formula.

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Questions students ask

What is the difference between nPr and nCr?

nPr counts ordered selections. nCr counts unordered groups and treats rearrangements of the same selected items as one result.

Why does the combination formula divide by r factorial?

The permutation count includes every ordering of the same r selected items. Dividing by r factorial removes those duplicates.

What is 0 factorial?

By definition, 0! equals 1. This keeps counting identities such as nC0 = 1 consistent.

Can I use nPr when repetition is allowed?

Not unchanged. For r ordered choices from n options with replacement, the count is n to the power r.