Permutations — When Order Matters

Suppose you're picking a president, vice president, and treasurer from a club of 10 people. Does it matter who gets which role? Absolutely — putting Maria as president and James as VP is completely different from putting James as president and Maria as VP. When the order of your choices matters, you're counting permutations.

This is the key idea: a permutation is an arrangement where order matters. Swap two things around and you get a different permutation.

Factorials First

Before getting to the formula, you need to understand factorials. The factorial of a number — written with an exclamation point — means multiply that number by every positive integer below it down to 1.

$$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$$ $$4! = 4 \times 3 \times 2 \times 1 = 24$$ $$3! = 6 \quad 2! = 2 \quad 1! = 1$$

And by convention, \(0! = 1\). That one surprises people, but it turns out to be necessary for the formulas to work cleanly.

Factorials grow fast. \(10!\) is already 3,628,800. This reflects just how many ways things can be arranged when you have a lot of items to work with.

When You Use All the Items

If you want to know how many ways you can arrange all of a group of items, it's just a factorial.

Example: In how many different orders can 4 runners finish a race (assuming no ties)?

The first place runner can be any of 4 people. Once that's settled, second place can be any of the remaining 3. Then 2 choices for third, and 1 person left for fourth.

$$4 \times 3 \times 2 \times 1 = 4! = 24$$

There are 24 possible orderings. Each one is a different permutation.

When You're Choosing a Subset

More often, you're picking a smaller group from a larger one — and order still matters. The formula for this is:

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

Where \(n\) is the total number of items and \(r\) is how many you're choosing. You'll also see this written as \(_nP_r\).

Example: A club has 8 members. They need to elect a president and a vice president. How many ways can this turn out?

Here \(n = 8\) and \(r = 2\).

$$P(8, 2) = \frac{8!}{(8-2)!} = \frac{8!}{6!} = \frac{8 \times 7 \times 6!}{6!} = 8 \times 7 = 56$$

Notice that the \(6!\) cancels out — you're really just multiplying the top two values: 8 and 7. That makes sense intuitively: 8 choices for president, then 7 remaining choices for VP.

Example: How many 3-letter arrangements can be made from the letters A, B, C, D, E (using each letter at most once)?

\(n = 5\), \(r = 3\)

$$P(5, 3) = \frac{5!}{(5-3)!} = \frac{5!}{2!} = \frac{120}{2} = 60$$

There are 60 different 3-letter arrangements. If this were a combination (where order didn't matter), the answer would be much smaller — but because ABC and BAC and CAB are all different arrangements, we're counting all of them.

Permutations vs. Combinations

This is the question that trips people up most often. Here's a quick way to think about it:

  • Permutation: You're picking people for specific roles. President vs. VP matters. First place vs. second place matters. The arrangement matters.
  • Combination: You're picking a group. A committee of 3 people is the same committee no matter what order you list them in. The arrangement doesn't matter.

Ask yourself: if I swap two of my choices, do I get something different? If yes — permutation. If no — combination.

Example: Picking a 4-digit PIN — permutation (1234 and 4321 are different PINs).

Example: Choosing 3 toppings for a pizza — combination (pepperoni + mushroom + olives is the same as olives + pepperoni + mushroom).

A Bigger Example

Example: A playlist has 10 songs. You want to pick 4 to play first, and the order matters (since you're cueing them up in sequence). How many ways can you do this?

$$P(10, 4) = \frac{10!}{6!} = 10 \times 9 \times 8 \times 7 = 5040$$

There are 5,040 possible 4-song opening sequences. That's a lot of variety for just 10 songs.

Try These

  1. In how many ways can 6 different books be arranged on a shelf? Show answer720 — \(6! = 720\)

  2. A race has 10 runners. How many different ways can first, second, and third place be awarded? Show answer720 — \(P(10, 3) = 10 \times 9 \times 8 = 720\)

  3. How many 4-digit codes can be formed from the digits 1–9 if no digit can be repeated? Show answer3,024 — \(P(9, 4) = 9 \times 8 \times 7 \times 6 = 3024\)

  4. A committee needs a chair, secretary, and treasurer. There are 12 volunteers. How many ways can the three roles be filled? Show answer1,320 — \(P(12, 3) = 12 \times 11 \times 10 = 1320\)

  5. Is choosing a 5-card hand from a deck of cards a permutation or combination? Explain. Show answerCombination — a hand of cards is the same hand regardless of what order you were dealt them. {A, K, Q, J, 10} is the same hand no matter how you arrange it.

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