Case 1
- Total items (n)
- 10
- Items arranged (r)
- 3
720
Open with these valuesPermutations Calculator
720permutations
Result: 720 permutationsA permutation counts orders, not just selections: Alice first, Bob second, Carol third differs from Carol, Alice, Bob. P(10, 3) = 10 × 9 × 8 = 720 podiums from ten runners. Ignore the order and the same question has only 120 answers — that is the combinations calculator.
The numbers at a glance
Held fixed: Total items (n) 10.
| Items arranged (r) | Result |
|---|---|
| 1 | 10 |
| 2 | 90 |
| 3Your value | 720 |
| 4 | 5,040 |
| 5 | 30,240 |
Worked examples
Case 2
- Total items (n)
- 52
- Items arranged (r)
- 5
311,875,200
Open with these valuesCase 3
- Total items (n)
- 8
- Items arranged (r)
- 3
336
Open with these valuesHow it's calculated
P(n, r) = n! ÷ (n − r)!
- StepEnter the size of the whole set as n — ten runners, 26 letters, 52 cards.
- StepEnter how many places you fill as r; each item may be used only once.
- ResultRead the number of ordered arrangements. Asking for more than exists gives 0.
What this number means
A permutation counts arrangements, and that count is a chain of shrinking choices. This is the whole idea behind n! ÷ (n − r)!, which looks stranger than what it does. Fill the places one after another: the first has all n available, the second only n − 1 because one of them is already standing there, the third n − 2. With ten runners and three podium places that is 10 × 9 × 8 = 720. The factorial form writes the same product as a fraction — n! runs the shrinking chain all the way down to 1, and (n − r)! is precisely the tail you never reach, so the division cancels it. The special cases fall out of the same picture instead of being rules of their own: at r = n the tail is 0! = 1 and the full n! remains, which makes P(5, 5) = 120, while at r = 0 the empty product is 1. What the formula assumes is that the n items are all different. Two identical items make it count arrangements nobody could tell apart, and the answer comes out too high — a word with a repeated letter is the everyday case. And all it counts is how many orders exist, never which of them is likely.
Order counts, and each item appears once
Every place is filled by a different item, and swapping two of them makes a new result. A podium from ten runners is the standard case: P(10, 3) = 10 × 9 × 8 = 720, and nobody takes two places.
Multiply r descending factors, not two factorials
Ten factorial divided by seven factorial cancels everything below 8, which leaves 10 × 9 × 8 = 720. The calculator multiplies exactly that product, so no oversized intermediate value ever appears.
The number of places stops at 11
P(55, 12) is the first result a standard number can no longer write out digit for digit: it prints as 210,152,248,122,136,300,000 when the true value ends in 320,000. Inside the allowed range every printed digit is true.
Commonly misread
A podium of three from ten runners has 120 outcomes.
That is the count with the order ignored. A podium ranks its three places, and each group of three comes in 3 factorial = 6 orders, so there are 720.
P(n, 0) is 0, because no place is filled.
It is 1, the single empty arrangement. P(n, n) is n factorial, the whole set put in order, so P(5, 5) is 120.
Asking for more places than runners just uses a runner twice.
It gives 0: with r greater than n no such arrangement exists. Reusing an item is a different question and a different calculator.
Reference table
| n, r | Reads as | P(n, r) |
|---|---|---|
| 0, 0 | arrange nothing at all | 1 |
| 5, 0 | the empty arrangement | 1 |
| 10, 1 | pick one of ten | 10 |
| 6, 2 | first and second from six | 30 |
| 8, 3 | a podium from eight runners | 336 |
| 10, 3 | a podium from ten runners | 720 |
| 5, 5 | order the whole set | 120 |
| 52, 5 | five cards dealt in order | 311875200 |
Questions
What is a permutation?
A permutation is an arrangement in which the order matters, so Alice, Bob, Carol is a different result from Carol, Alice, Bob. The count is written P(n, r), spoken n permute r, and equals n!/(n−r)!. Arranging 3 out of 10 gives 720.
How are permutations different from combinations?
Permutations count arrangements where order matters; combinations count selections where it does not. Each selection of r items can be arranged in r factorial ways, so permutations are always the larger count. Three from ten gives 720 permutations but only 120 combinations.
How do I calculate P(10, 3) by hand?
Multiply the r descending factors starting at n: 10 × 9 × 8 = 720. The full factorials are never needed, because dividing 10 factorial by 7 factorial cancels everything below 8. This calculator uses exactly that product, so no oversized intermediate value ever appears.
What are P(n, 0) and P(n, n)?
P(n, 0) is 1, the single empty arrangement. P(n, n) is n factorial, the whole set put in order, so P(5, 5) is 120. Asking for more places than items, with r greater than n, gives 0, because no such arrangement exists.
Why does r stop at 11?
Because P(55, 12) is the first result a standard number can no longer write out digit for digit: it prints as 210,152,248,122,136,300,000 when the true value ends in 320,000. Every answer inside the allowed range is exact to the last digit. Larger arrangements need arbitrary-precision arithmetic.
Sources and last check
Information, not professional advice.