Permutations vs Combinations: When Order Matters

Counting & combinatorics · Updated July 2026

Counting problems have a special way of humiliating smart students. The arithmetic is trivial — multiply a few numbers — but choosing which numbers to multiply requires reading the problem like a lawyer. The good news: the vast majority of introductory counting problems are settled by a single question. This guide gives you that question, the formulas it selects between, and the standard traps.

The one question: does order matter?

Whenever a problem asks "how many ways can we pick r things from n things," stop and ask: if I picked the same items in a different order, would that count as a different outcome?

• Yes → permutation. Passwords, rankings, race podiums, seating arrangements. Alice-then-Bob differs from Bob-then-Alice.
• No → combination. Committees, teams, hands of cards, pizza toppings. A committee of Alice and Bob is the committee of Bob and Alice.

A quick vocabulary test: "arrangement," "sequence," "order," and "ranking" signal permutations; "selection," "subset," "group," and "committee" signal combinations. But do not trust keywords alone — reframe the outcome and ask whether swapping two chosen items changes it.

The formulas, and where they come from

Everything starts with the product rule: if a process has k independent stages with n₁, n₂, …, n_k options, the total number of outcomes is the product. The factorial n! = n × (n−1) × … × 1 counts full arrangements of n distinct items: n choices for the first slot, n−1 for the second, and so on.

Permutations of r from n:

P(n, r) = n! / (n − r)! — fill r ordered slots: n options, then n−1, …, down to n−r+1.

Combinations of r from n:

C(n, r) = n! / (r!(n − r)!) — take the ordered count P(n, r) and divide out the r! orderings of each chosen set, since order no longer distinguishes outcomes.

That derivation is worth remembering, because "divide by what you overcounted" is the master move of combinatorics. C(n, r) is also called the binomial coefficient, written (n choose r), and it is exactly the coefficient of x^r in the expansion of (1+x)^n — the binomial theorem is a counting statement in algebraic clothing.

Worked contrast

From 10 students: how many ways to pick a president, secretary, and treasurer? Roles are distinguishable, so order matters: P(10, 3) = 10 × 9 × 8 = 720. How many ways to pick a 3-person committee? Order does not matter: C(10, 3) = 720 / 3! = 120. Same underlying choices; the factor of 6 is exactly the number of ways to shuffle three people among identical seats.

The pigeonhole principle

Counting's strange sibling deserves a mention because it shares the same exam. The pigeonhole principle says: if you place more than n items into n boxes, some box holds at least two. It sounds too obvious to be useful, yet it proves striking facts — among any 13 people, two share a birth month; in any group of people, two have shaken the same number of hands. The skill is choosing what the "boxes" are. When a problem says "show that at least two…" with no obvious structure, pigeonhole is the tool to reach for.

Stars and bars: counting with repetition

What if you distribute 10 identical candies among 4 kids? Items are identical and only the counts matter, so neither P nor C applies directly. The stars-and-bars trick represents each distribution as a row of 10 stars (candies) separated by 3 bars (dividers between kids): ★★★|★★|★★★★|★. Every arrangement of 10 stars and 3 bars is one distribution, so the answer is C(13, 3) = 286 — choose which 3 of the 13 symbols are bars. In general, distributing n identical items into k distinct bins gives C(n + k − 1, k − 1). This one idea dispatches a whole family of "how many solutions does x₁ + x₂ + … + x_k = n have" problems.

Traps that cost exam points

• Overcounting duplicates. If some items are identical (the letters of MISSISSIPPI), divide by the factorials of the repeat counts.
• Adding when you should multiply. Stages of one process multiply; mutually exclusive alternatives add. "And" is ×, "or" is + (when the cases don't overlap).
• Forgetting the complement. "At least one" is usually easiest as total minus "none."
• Mixing P and C in multi-part problems. A problem can need both: choose the committee (C), then arrange its officers (P). Handle each stage on its own terms, then multiply.

Sanity checks help: C(n, r) must equal C(n, n−r); P(n, r) must be at least C(n, r); and small cases can be brute-forced by listing. If your formula says 6 but you can only list 3 outcomes, the formula loses.