Arrangement without repetition: P(8,3)
- Input
- n=8, r=3
- Expected output
- P(8,3) = 336
336 is 6 times the equivalent combination C(8,3) = 56, because each chosen trio can be ordered in 3! = 6 different ways.
Carregando ferramenta… / Loading tool…
permutations and arrangements
Permutation counts ordered arrangements: how many different sequences can be built by choosing r items out of n total, with order taken into account. This page shows the formula, the special case where r equals n, and the case where repetition is allowed.
336 is 6 times the equivalent combination C(8,3) = 56, because each chosen trio can be ordered in 3! = 6 different ways.
When all n items enter the arrangement, the permutation collapses into the full factorial of n; there are 120 ways to arrange 5 distinct books in a row.
Each of the 4 positions can repeat any of the 10 digits independently, so the total is 10 multiplied by itself 4 times, well above the 5,040 codes with no repeated digit.
A permutation is an ordered arrangement of r elements chosen from a set of n; order matters, so AB and BA count as different permutations. Without repetition the number of arrangements is P(n, r) = n! / (n − r)!. For example, arranging 3 of the letters A, B, C, D gives P(4, 3) = 4 · 3 · 2 = 24 ordered sequences.
They are the same concept under different names: "arrangement" is the more common term in some curricula for P(n,r) when r is less than n, while "simple permutation" is often reserved for the r = n case, but both use the same n!/(n-r)! formula.
By mathematical convention: there is exactly one way to arrange zero items, the empty arrangement, so 0! = 1 keeps the formula P(n,n) = n!/0! consistent without needing a separate special case for when every item enters the arrangement.
10! = 3,628,800, computed as P(10,10); that factorial growth is fast, 10 items already exceed 3.6 million possible orderings, and 15 items already exceed 1.3 trillion.
Whenever the same item can occupy more than one position, like digits in a code, letters on a license plate or faces in a repeated dice roll; in those cases PR(n,r) = n^r is the right formula, not P(n,r), because the "no repeats" restriction does not exist in the original problem.
P(n, r) = n! / (n−r)!
All calculations stay in your browser. No data is sent to any server.