k-Permutation¶

Definition¶

Let $A$ be a finite set with $n$ elements and let $0\le k\le n$ $n$ . A $k$ -permutation of $A$ is an ordered selection $(a_1,\ldots,a_k)$ of $k$ distinct elements of $A$ . For $k=n$ it is an ordering of all of $A$ , which corresponds to a bijection $\{1,\ldots,n\}\to A$ $A$ ; when $A=\{1,\ldots,n\}$ this is a permutation of $A$ in the Math II sense of a bijection from $A$ onto itself.

Interpretation¶

A $k$ -permutation is an ordered selection: the same objects in a different order count as a different selection. It counts rankings, arrangements in a row and assignments of objects to labelled positions. Counting texts often say permutation for both the full arrangement and the partial selection; the $k$ makes clear which one is meant.

All six arrangements of the letters A, B and C in a row.

The figure lists all $3!=6$ arrangements of $A$ , $B$ and $C$ . They are found by filling the positions from left to right: $3$ choices for the first letter, $2$ for the second and $1$ for the last, so $3\cdot2\cdot1=6$ by the multiplication rule.

The same reasoning for $k$ positions and $n$ objects gives $n(n-1)\cdots(n-k+1)$ $1)$ : each factor is one smaller than the previous one because an object that has been used cannot be chosen again. For a podium of three among eight runners this is $8\cdot7\cdot6=336$ $336$ . The form $n!/(n-k)!$ gives $n!/0!=n!$ for $k=n$ $n$ , which is one reason for the convention $0!=1$ $1$ . Dividing by the $k!$ orders of each group turns the count into the number of combinations.

A common slip is to use this count when repetition is allowed. Four-digit PIN codes may repeat digits, so there are $10^4=10\,000$ of them, not $10\cdot9\cdot8\cdot7=5040$ $5040$ ; see sampling with and without replacement. The full case $k=n$ with $A=\{1,\ldots,n\}$ is what Math II calls a permutation.

Key Results¶

The number of $k$ -permutations of $n$ distinct objects is $n(n-1)\cdots(n-k+1)=n!/(n-k)!$ $k)!$ .
In particular there are $n!$ orderings of $n$ distinct objects.

Examples¶

The number of possible podium finishes among eight runners is a permutation count.
Arranging five books on a shelf uses full permutations.
Assigning three distinct tasks to three ordered time slots is a permutation problem.

Prerequisites¶

Set

Sources¶

Probability Theory for EOR · Week 1

Aliases: partial permutation, ordered selection, variation Keywords: arrangement, permutation

Last updated: 20 October 2026 03:02