Statistics › 🎲 Probability

Counting: Permutations & Combinations

How to count arrangements and selections without ever writing them all out — and how to tell which one a problem is asking for.

Intro StatisticsAP Statistics level
💡
The big idea: Probability almost always starts with a count: how many ways can this happen? Instead of listing every possibility by hand, a few counting rules do the work. The multiplication principle chains choices together, factorials count full arrangements, and two close cousins — permutations and combinations — handle the everyday question of picking some items out of many. The only thing you ever have to decide is whether the order of what you picked matters.
🎯 By the end, you'll be able to
  • Use the multiplication principle to count the outcomes of a sequence of independent choices
  • Compute factorials and use them to count full arrangements of a set
  • Tell permutations (order matters) apart from combinations (order ignored) and pick the right one for a problem
  • Move between the two counts using the relationship nPr = nCr times r!
📎 You should already know
  • Basic probability and the idea of an outcome
  • Comfort with multiplication and exponents

Counting without listing everything

Suppose a lunch deal lets you pick one of 3 sandwiches, one of 4 sides, and one of 2 drinks. How many different lunches are possible? You could try to list them all — but there is a faster way. You have 3 choices, then for each of those 4 more, then for each of those 2 more, so the choices multiply: \(3\times4\times2=24\).

That simple habit — multiply the number of options at each step — is the engine behind almost every counting problem in probability. Everything else in this lesson is just careful bookkeeping on top of it.

🔑 The multiplication principle
If one choice can be made in a ways, and a second independent choice in b ways, then the two choices together can be made in a × b ways. This extends to any number of steps: multiply the options at every step to get the total number of outcomes.

Factorials: arranging a whole set

Now ask a sharper question: in how many orders can you line up all the items in a set? For 3 books on a shelf, there are 3 choices for the first slot, 2 left for the second, and 1 for the last: \(3\times2\times1=6\). This running product of an integer down to 1 is so common it gets its own symbol, the factorial, written with an exclamation mark.

So \(3!=6\), \(4!=24\), and factorials grow fast — \(10!\) is already over three million. By convention \(0!=1\) (there is exactly one way to arrange nothing).

\[ n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1 \]
The factorial of n: the number of ways to arrange all n distinct items in a row. Read n! as 'n factorial'.

Order matters — or it doesn't

Usually you do not arrange the whole set; you pick \(r\) items out of \(n\). Here everything hinges on a single question: does the order of the picks matter?

  • If order matters — like awarding 1st, 2nd, and 3rd place — each ordering is a different outcome. These are permutations, counted by \({}^{n}P_{r}\).
  • If order is ignored — like choosing a committee, where the same people are the same group no matter who was named first — each unordered group counts once. These are combinations, counted by \({}^{n}C_{r}\).

A permutation cares about arrangement; a combination cares only about which items got picked.

\[ {}^{n}P_{r} = \frac{n!}{(n-r)!} \qquad\qquad {}^{n}C_{r} = \frac{n!}{r!\,(n-r)!} \]
Permutations count ordered selections of r from n; combinations count unordered selections. The extra r! in the denominator of nCr divides out the orderings that permutations counted separately.
🎮 Interactive: order matters vs order ignored LIVE
Set the pool size n and how many you choose, r. The tool computes the factorial, the number of ordered arrangements (nPr), and the number of unordered selections (nCr), and lays the two side by side so you can see the same r items counted as ordered slots versus one unordered set.
✨ Why nCr is always smaller — and by exactly how much
Every unordered selection of \(r\) items can itself be shuffled into \(r!\) different orders. So permutations overcount combinations by exactly a factor of \(r!\): first pick the group (\({}^{n}C_{r}\) ways), then arrange it (\(r!\) ways). That is why dividing \({}^{n}P_{r}\) by \(r!\) turns arrangements back into plain selections.
\[ {}^{n}P_{r} = {}^{n}C_{r} \times r! \]
The bridge between the two: arrangements equal selections times the number of ways to order each selection.

Which one should I use?

When a problem asks 'how many ways…', pause and ask whether swapping the order of the chosen items gives a genuinely different result:

  • Different result → permutations. Rankings, finishing positions, passwords and PINs, seating people in a row, assigning distinct job titles.
  • Same result → combinations. Choosing a committee, dealing a hand of cards, picking pizza toppings, selecting a subset with no roles or ranks.

A quick tell: if the picked items get labels, seats, or ranks, order matters. If they just go into an unlabeled group, it does not.

📝 Worked example: A club has 10 members. (a) In how many ways can it elect a president, a vice-president, and a treasurer? (b) In how many ways can it instead form a 3-person committee with no titles?
  1. Part (a) — the three roles are distinct, so order matters: this is a permutation. \({}^{10}P_{3} = 10\times9\times8 = 720\).
  2. Part (b) — a committee has no roles, so the same three people are the same committee in any order: this is a combination. \({}^{10}C_{3} = \frac{720}{3!} = \frac{720}{6} = 120\).
  3. Sanity check with the bridge: \({}^{n}P_{r} = {}^{n}C_{r}\times r!\) gives \(720 = 120\times 6\). Each committee of 3 can be arranged into \(3!=6\) different slates of officers.
✓ There are 720 ways to fill the three distinct offices (permutations) but only 120 possible committees (combinations) — exactly 6 times fewer, because each committee corresponds to 3! = 6 different orderings.

Check your understanding

1. A café offers 3 sizes, 4 flavors, and 2 milk options. Using the multiplication principle, how many different drinks can you order?
Multiply the options at each step: 3 × 4 × 2 = 24 different drinks.
2. You will place 3 of your 5 trophies in a row on a shelf, and the order on the shelf matters. How many arrangements are possible?
Order matters, so this is a permutation: 5P3 = 5 × 4 × 3 = 60.
3. From the same 5 trophies you instead just choose 3 to donate, with no ordering. How many selections are possible?
Order is ignored, so this is a combination: 5C3 = 5P3 / 3! = 60 / 6 = 10.
4. Which of these situations is counted with a COMBINATION (order does not matter)?
Toppings go into an unlabeled group, so order does not matter — a combination. Rankings, passwords, and seating all depend on order, so they are permutations.
✅ Key takeaways
  • The multiplication principle: multiply the number of options at each step to count the outcomes of a sequence of choices.
  • A factorial n! counts the ways to arrange all n distinct items; factorials grow very fast and 0! = 1 by convention.
  • Permutations (nPr = n!/(n-r)!) count ordered selections — use them when order matters, like rankings or seats.
  • Combinations (nCr = n!/(r!(n-r)!)) count unordered selections — use them when order is ignored, like committees or hands of cards.
  • The two are linked by nPr = nCr times r!: each selection of r items can be arranged in r! orders, so permutations overcount combinations by exactly r!.