Permutations Calculator (nPr)

Calculate P(n,r), the number of ordered arrangements of r items chosen from n, with the n!/(n−r)! formula, factorial cancellation, and step-by-step work shown.

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How to use this calculator

Enter n (the total number of distinct items available) and r (how many you are arranging). The calculator returns P(n, r) — the number of ordered arrangements — along with the factorial formula, the cancelled-down multiplication chain, and a full step-by-step breakdown.

Tip: Permutations answer "how many ways can I arrange r items in order?" If the order of the outcome doesn't matter, you want a combination (nCr) instead — see the comparison below.

Order matters — the defining idea

A permutation counts ordered arrangements. Picture a race with 10 runners where you want to know the number of possible 1st/2nd/3rd place outcomes. Swapping who comes in 1st and who comes in 3rd produces a different result, even though the same three runners are on the podium. That sensitivity to order is exactly what nPrnPr measures.

P(n,r)=n!(nr)!P(n,r) = \frac{n!}{(n-r)!}

Worked example — P(10,3)

P(10,3)=10!7!=10×9×8=720P(10,3) = \frac{10!}{7!} = 10 \times 9 \times 8 = 720

The factorial ratio 10!/7!10!/7! looks intimidating until you notice that 7!7! cancels almost all of 10!10! — everything from 7 down to 1 divides out, leaving only the top three factors: 10×9×810 \times 9 \times 8. That cancellation is why the calculator's expansion line shows a short product instead of two huge factorials: for any P(n,r)P(n,r), only rr factors ever survive, counting down from nn.

Permutations vs. combinations

Both start from the same nn items, but they answer different questions:

Permutation nPrCombination nCr
QuestionIn how many orders can I arrange r items?In how many ways can I select r items?
Formulan!/(nr)!n!/(n-r)!n!/(r!(nr)!)n!/(r!(n-r)!)
Example (10 people, r = 3)President/VP/Secretary: 720 ways3-person committee: 120 ways
RelationshipnCr=nPr/r!nCr = nPr / r!

The relationship is direct: a combination is a permutation with the r!r! internal orderings of each group collapsed into one. C(10,3)=P(10,3)/3!=720/6=120C(10,3) = P(10,3)/3! = 720/6 = 120.

The classic permutation-vs-combination mistake

The single most common error in counting problems is picking the wrong formula because the "order matters?" question was never asked. Two people, two prizes:

  • First prize and second prize are different awards → order matters → use nPrnPr.
  • Both winners get an identical gift card → order doesn't matter → use nCrnCr.

Same 2 people chosen from the same pool, but the count is different: P(n,2)P(n,2) counts each pair twice (once per ordering), while C(n,2)C(n,2) counts it once. Always ask whether swapping two chosen items changes the outcome before reaching for a formula.

Special cases

  • P(n, 0) = 1 — there is exactly one way to arrange zero items: do nothing (the empty arrangement).
  • P(n, n) = n! — when every item gets a position, the permutation formula reduces to a plain factorial, since (nn)!=0!=1(n-n)! = 0! = 1.
  • P(n, 1) = n — arranging a single item chosen from n is just picking which one.

Frequently asked questions

What is P(10,3)?
P(10,3) counts the ordered arrangements of 3 items chosen from 10. Using P(n,r)=n!/(nr)!P(n,r) = n!/(n-r)!: 10!/7!=10×9×8=72010!/7! = 10 \times 9 \times 8 = 720.
What does nPr mean?
nPr — written P(n,r)P(n,r) — is the number of ways to arrange rr items chosen from a set of nn distinct items where the order of the arrangement matters. First place, second place, and third place are different outcomes even with the same three people.
How is a permutation different from a combination?
A permutation counts ordered arrangements; a combination counts unordered selections. Choosing a president, VP, and secretary from 10 people is a permutation (720 ways) because the roles differ; choosing an unordered 3-person committee from the same 10 people is a combination (120 ways) because no one has a distinct role.
What is the most common mistake with permutations?
Confusing permutations with combinations — using nPrnPr when order does not actually matter, or nCrnCr when it does. Ask first: does swapping two chosen items produce a different outcome? If yes, use nPrnPr; if no, use nCrnCr.
Why does P(n,0) always equal 1?
Choosing 0 items has exactly one outcome — the empty arrangement — so P(n,0)=1P(n,0) = 1 for any nn. This matches the formula: n!/(n0)!=n!/n!=1n!/(n-0)! = n!/n! = 1.
What is P(n,n)?
When r=nr = n, every item gets placed, so P(n,n)=n!P(n,n) = n!. For example P(5,5)=5!=120P(5,5) = 5! = 120 — the number of ways to line up all 5 items.
How does nPr relate to nCr?
nCr=nPr/r!nCr = nPr / r!. A permutation counts every ordering of the same rr items separately, while a combination treats those r!r! orderings as one outcome, so dividing by r!r! converts arrangements into selections. C(10,3)=720/6=120C(10,3) = 720/6 = 120.

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