1
Each choice removes one option
n ways for the first position, n−1 for the second, and so on for r positions. That product is P(n, r).
Lesson · Combinatorics and Counting · Arrangements
Permutations
P(n, r) = n!/(n−r)! counts the ways to arrange r of n distinct objects in order. Factorials grow explosively, and that growth is a fact about the problem rather than about the notation.
P(n, r) = n(n−1)⋯(n−r+1) = n!/(n−r)!
P(n, n) = n! all n objects arranged
0! = 1 the empty arrangement
circular arrangements of n: (n−1)!
Ordered, no repetition
Core ideas
What this lesson establishes
The product rule with a shrinking pool of choices. Everything else is notation.
2
0! = 1 is a convention that works
It makes P(n, n) = n!/0! = n! come out right and keeps every formula in this course free of special cases.
3
Factorial growth is genuinely enormous
52! is about 8 × 10⁶⁷. Every shuffle of a deck that has ever been performed was almost certainly unique.
The material
Arrangements
The formula, its derivation, and two variants worth knowing.
Definition · Permutation
A permutation of a set is an ordered arrangement of its elements. An r-permutation is an ordered arrangement of r of them.
Theorem · Counting r-permutations
For 0 ≤ r ≤ n, P(n, r) = n(n−1)(n−2)⋯(n−r+1) = n!/(n−r)!.
Fill the r positions in order. The first has n candidates. Whatever was chosen, the second has n−1 remaining, the third n−2, and the rth has n−r+1. The number of options at each stage does not depend on which particular objects were chosen earlier, so the product rule applies.
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Note the count of factors: r of them, running from n down to n−r+1. Writing n down to n−r is the standard off-by-one, and it gives P(n, r+1).
Example · Podium finishes
Ten runners, how many ways to fill gold, silver and bronze? P(10,3) = 10 · 9 · 8 = 720. Order matters here — the same three runners in a different order is a different outcome — which is exactly what makes it a permutation rather than a combination.
Circular arrangements
Theorem · Arrangements around a circle
n distinct objects can be arranged in a circle in (n−1)! ways, when rotations are considered identical.
There are n! linear arrangements. Each circular arrangement corresponds to exactly n of them, one for each choice of which object is written first. So the circular count is n!/n = (n−1)!.
Equivalently: fix one object's position to break the rotational symmetry, and arrange the remaining n−1 freely.
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The division is legitimate only because rotations were declared identical. If the chairs are numbered, rotating everyone one seat along produces a different seating, nothing is identified, and the count is n!. Say which situation you are in before dividing; the question decides it, not the shape of the table.
The division-by-symmetry argument is the pattern to remember: when several arrangements are considered the same, count them all and divide by the size of the equivalence class. That is Sets, Relations, and Functions's equivalence relation doing combinatorial work, and “Permutations with Repetition” uses it again.
The size of factorials
5! = 120
10! = 3 628 800
20! = 2 432 902 008 176 640 000
52! ≈ 8.07 × 10⁶⁷
52! is about 10¹⁸ times the number of atoms in the Earth, which is roughly 10⁵⁰. Any algorithm that examines all permutations of a modest input is not slow — it is impossible, and “P, NP and NP-Completeness” in Algorithms and Complexity gives that observation a name.
Interactive
The four counting rules
Enumerated while enumeration is possible
Method
Counting arrangements
Confirm order matters before reaching for the formula.
1
Check that order matters
If rearranging the same objects gives a different outcome, it is a permutation. If not, “Combinations” has the right formula.
2
Check that repetition is forbidden
Objects are drawn without replacement. If they may repeat, “Permutations with Repetition” applies and the answer is nʳ.
3
Count the factors
P(n, r) has exactly r factors, ending at n−r+1. Miscounting them is the standard error.
4
Divide out any symmetry
Circular seating divides by n; a necklace that may be flipped divides by 2n. Name the symmetry before dividing.
Permutations
Score 0 / 0Choose the answer you can justify.
Seating with a constraint
Eight people at a round table; two of them must sit together.
Unrestricted circular arrangements: (8−1)! = 5040
Treat the pair as a single block:
7 items around a circle: (7−1)! = 720
the pair can be in 2 internal orders: × 2
⟹ 1440 arrangements with the two together
With the two apart: 5040 − 1440 = 3600
Check: 1440 / 5040 = 2/7. Fix one of the pair; the other has
7 seats to choose from, and exactly 2 of them are beside the first.
The block trick — glue constrained objects together, count, then multiply by their internal arrangements — is the standard device for "must be adjacent" constraints. The final ratio check is cheap and catches an error in the block count immediately.
Common mistakes
1
Using P when order does not matterA committee of three is not an ordered triple. P(n,r) counts each committee r! times, which is exactly the relationship “Combinations” exploits.
2
Miscounting the factorsP(n, r) runs from n down to n − r + 1, which is r factors. Ending at n − r gives one too many.
3
Forgetting to divide out a symmetryCircular arrangements, necklaces and identical items all require dividing by the size of the symmetry class. Naming the symmetry first prevents dividing by the wrong number.
Check your understanding
✓
Finish when you can justify the divisor in a symmetry argument.
Count the ways to seat 6 people around a round table with two specific people not adjacent. Do it by complement, and state exactly why the circular count is 5! rather than 6!.
Note: A permutation is also a bijection from a set to itself, which is the remark of “Composition and Inverse Functions”, in Sets, Relations, and Functions. There are n! of them, and under composition they form the symmetric group — the object most of finite group theory is about.