Counting: Permutations & Combinations
You'll learn to count equally-likely outcomes with the multiplication principle, and tell permutations (order matters) from combinations (order doesn't) so you can plug the count into \(P(\text{event}) = \text{favorable}/\text{total}\).
Predict: if you bump k from 3 to 4 (leaving n fixed), will the gap between P(n,k) and C(n,k) grow or shrink? Then try it below to check. Pick how many items you have (n) and how many you're choosing (k) — watch the ordered count P(n,k) and unordered count C(n,k) diverge. P always overcounts C by a factor of k! (every unordered group can be arranged k! different ways). Hover a bar for the exact value.
\(P(n,k) = \dfrac{n!}{(n-k)!}\)
\(C(n,k) = \dfrac{n!}{k!(n-k)!}\)
Counting techniques enumerate equally likely outcomes so you can compute \(P(\text{event}) = \text{favorable} / \text{total}\).
Say you're arranging books on a shelf — order matters, so that's a permutation. Now say you're just picking a 3-person committee (nobody has a "role") — order doesn't matter, so that's a combination. Combinations are always smaller (or equal), because they collapse every re-ordering of the same group into one count.
The multiplication principle: you get \(n_1 \times n_2 \times \cdots \times n_k\) ways to complete \(k\) independent steps. A factorial \(n!\) counts full orderings. Permutations, from sampling without replacement when order matters: \(P(n,k) = n!/(n-k)!\). Combinations, when order doesn't matter: \(C(n,k) = n!/(k!(n-k)!)\), and \(C(n,k) = C(n, n-k)\).
Imagine you're an actuary selecting a random sample of 5 claims from a batch of 52 files for audit. You'd use \(C(52,5)\) to count how many distinct samples are possible — the basis for computing the probability that a specific subset of "flagged" claims appears in the audit sample (a hypergeometric setup common on Exam P).
A lottery draws 3 winning numbers from 25 (order doesn't matter).
Step 1 — which tool applies? Combination, since order doesn't matter.
Step 2 — set up the formula: \(C(25,3) = \dfrac{25!}{3! \cdot 22!}\).
Step 3 — compute the value:
Reveal the answer
More info — why dividing by k! removes duplicate orderings
Here's another way to see \(C(n,k) = P(n,k)/k!\): first pick k items in order — that's \(P(n,k)\) ways. Now notice every unordered group of k items got counted once for each of the k! ways to order that same group. Dividing by k! collapses those duplicates back down to one count per group. See the Khan Academy counting unit under Dive deeper below for graded practice on this.
Check your understanding
How many distinct 5-card hands can be dealt from a standard 52-card deck, where the order the cards are received doesn't matter?
A committee chooses 3 officers (president, secretary, treasurer — distinct roles) from 8 candidates. How many ways can this be done?
When k = n (you're choosing or arranging every item), what's true about P(n,n) and C(n,n)?
A batch of 10 files contains 3 flagged for review. You randomly select a sample of 4 files for audit (order doesn't matter). What's P(at least one flagged file is in the sample)?
Recap
- Multiplication principle: \(n_1 \times n_2 \times \cdots \times n_k\) ways to complete k independent steps.
- Permutation \(P(n,k) = n!/(n-k)!\) — order matters.
- Combination \(C(n,k) = n!/(k!(n-k)!)\) — order doesn't matter; always ≤ P(n,k), off by exactly a factor of k!.
- Plug the count into \(P(\text{event}) = \text{favorable}/\text{total}\).
Dive deeper
- Khan Academy — Counting, permutations, and combinations Build permutation and combination fluency with graded practice.
- Seeing Theory — Compound Probability Visualize how ordered and unordered counts differ before the formulas.
- Harvard Stat 110 — Lecture 1: Probability and Counting A rigorous first-principles derivation of the counting rules.
Sources
- Combinatorics — Counting for Probability