Permutations | Moosa Academy

Permutations — Moosa Academy

A permutation is an arrangement where order matters. Unlike a combination, ABC and BAC count as two different permutations — from the three letters $\{A, B, C\}$ there is one combination but six permutations.

Permutations Arrangements, in order

A permutation chooses $r$ elements from $n$ with regard to order. The same elements in a new order make a new permutation.

From $\{A, B, C\}$ , the arrangements of all three are ABC, ACB, BAC, BCA, CAB, CBA — six permutations, but only one combination $\{A, B, C\}$ .

Formula The three laws

Without repetition (each element used once):

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

With repetition (an element may be reused):

P(n, r) = n^{r}

Arranging all

n

(when

r = n

P(n, n) = n!

Allowing repetition always gives a larger count, since no choice is ever used up.

Example Filling the slots

A 4-digit job number from the digits 1–9, no digit repeated. Each slot has one fewer choice than the last:

P(9, 4) = \dfrac{9!}{5!} = 9 \times 8 \times 7 \times 6

⟹ 3,024 numbers.

Example A few more

Arranging 3 books chosen from 7

P(7, 3) = 7 \times 6 \times 5

⟹ 210 arrangements.

A 3-letter password from {A, B, C, D, E}, no repeats

P(5, 3) = 5 \times 4 \times 3

⟹ 60 passwords.

A 5-digit security code from 0–9, repetition allowed

Every position has all 10 digits:

10^{5}

⟹ 100,000 codes.

Arranging all 5 letters of a word (all different)

P(5, 5) = 5!

⟹ 120 arrangements.

Reference The three types

Type	Formula	Example	Result
No repetition	n! ÷ (n − r)!	P(9, 4)	3,024
With repetition	nʳ	9⁴	6,561
Complete (r = n)	n!	P(5, 5)	120

Summary

Permutations arrange $r$ of $n$ where order matters: ABC ≠ BAC.
Without repetition: $P(n, r) = \dfrac{n!}{(n-r)!}$ — each element is used once.
With repetition: $P(n, r) = n^{r}$ — always a larger count.
Arranging all $n$ elements (when $r = n$ ) gives $n!$ .