Permutations: Arrangements Where Order Matters

What is a permutation?

A permutation is an ordered arrangement of objects. If you care who comes first, second, third, you are counting permutations. Rearranging the same group into a different order counts as a different permutation.

The symbol $P(n, r)$ — sometimes written $_nP_r$ — means "pick $r$ items from a pool of $n$ and put them in order."

The formula

$P(n, r) = \dfrac{n!}{(n - r)!} = n \cdot (n - 1) \cdot (n - 2) \cdots (n - r + 1)$

The expanded product has exactly $r$ factors, starting at $n$ and counting down.

Step-by-step solution

Five runners compete. We want to know the number of ways to assign gold, silver, and bronze (so $n = 5$, $r = 3$).

Step 1 — Choose 1st place: Any of the 5 runners could win. 5 choices.

Step 2 — Choose 2nd place: One runner is already 1st, so 4 remain. 4 choices.

Step 3 — Choose 3rd place: Two are already placed. 3 choices.

Step 4 — Multiply (counting principle): $P(5, 3) = 5 \cdot 4 \cdot 3 = \boxed{60}$

Step 5 — Verify with the factorial formula: $P(5,3) = \dfrac{5!}{(5-3)!} = \dfrac{5!}{2!} = \dfrac{120}{2} = 60 \checkmark$

Why order matters here

Runner Alice finishing 1st with Bob 2nd and Carol 3rd is a different medal ceremony than Bob 1st, Alice 2nd, Carol 3rd. The same three people, different arrangement → counted separately.

Contrast: if the problem said "choose 3 runners for the team" (no ranks), then order would not matter and we would use combinations instead.

Special case: arranging all $n$ items

When $r = n$, the formula gives $P(n, n) = n!$. For example, the number of ways to line up all 5 runners is $5! = 120$.

Common mistakes

• Forgetting that order matters. If you accidentally use $C(n, r)$ (combinations), you undercount by a factor of $r!$.
• Using the wrong number of factors. The expanded product $n(n-1)(n-2)\cdots$ has exactly $r$ terms, not $n$. For $P(5, 3)$ it is three factors, not five.
• Mixing up repetition rules. The formula $n!/(n-r)!$ assumes no repeats. If positions can repeat (like a 4-digit PIN where digits can repeat), it is $n^r$ instead.

Try it in the visualization

Move the $n$ and $r$ sliders and watch the branching tree fill in. Each path from root to leaf is one permutation; the count at the bottom is $P(n, r)$.