Arrangements: Permutations of $n$ Objects Taken $k$ at a Time
Also called $k$-permutations, denoted $P(n,k)$ or $A(n,k)$
Prerequisite: You should already be comfortable with factorials ($n!$) and full permutations of $n$ objects. This lesson extends those ideas to selecting and ordering only $k$ of the $n$ objects.
1. What Is an Ordered Selection Without Repetition?
Imagine you have $n$ distinct objects and you want to choose $k$ of them, where the order in which you choose them matters and no object can be chosen twice. Each such ordered selection is called a $k$-permutation (or arrangement) of $n$ objects.
Key Idea: A $k$-permutation counts two things simultaneously — which $k$ objects you pick and in what order you place them.
Everyday Examples
| Scenario | $n$ | $k$ | Why Order Matters |
|---|---|---|---|
| Awarding gold, silver, and bronze medals to 3 of 10 athletes | 10 | 3 | Gold–Silver–Bronze is different from Silver–Gold–Bronze |
| Choosing a president, vice-president, and treasurer from 8 club members | 8 | 3 | Each position is distinct |
| Forming a 4-letter "word" from 26 distinct letters (no repeats) | 26 | 4 | ABCD ≠ DCBA |
| Seating 5 people in 3 numbered chairs | 5 | 3 | Chair 1 vs Chair 2 vs Chair 3 are different positions |
2. Deriving the Formula via the Multiplication Principle
We fill the $k$ positions one at a time:
| Position | Available choices |
|---|---|
| 1st | $n$ |
| 2nd | $n-1$ |
| 3rd | $n-2$ |
| $\vdots$ | $\vdots$ |
| $k$-th | $n - k + 1$ |
By the multiplication (product) rule:
$$ P(n,k) = n \cdot (n-1) \cdot (n-2) \cdots (n-k+1) $$This is a product of exactly $k$ consecutive descending integers starting from $n$.
Compact Factorial Form
Multiply and divide by $(n-k)!$:
$$ P(n,k) = \frac{n!}{(n-k)!} $$
Why does this work? The numerator $n!$ contains all factors from $n$ down to $1$. Dividing by $(n-k)!$ cancels the factors $(n-k)(n-k-1)\cdots 2 \cdot 1$, leaving exactly the $k$ factors $n(n-1)\cdots(n-k+1)$.
Quick Derivation Diagram
For $P(5,3)$:
$$ P(5,3) = 5 \times 4 \times 3 = 60 \quad\text{or equivalently}\quad \frac{5!}{(5-3)!} = \frac{120}{2} = 60 $$