Every statistics textbook presents a mnemonic for permutations versus combinations, and virtually every student forgets it mid-exam. The reason mnemonics fail here is that the distinction does not lie in the formula — it lies in a question about the counting problem itself. Once you can answer that question, the formula is irrelevant.
The one question
Does the order of selection matter?
If you are choosing 3 people from a group of 10 to be President, Vice-President, and Treasurer, the order matters. Alice as President is a different outcome from Alice as Treasurer. This is a permutation.
If you are choosing 3 people from a group of 10 to form a committee, the order does not matter. A committee of Alice, Bob, and Carol is the same committee regardless of which name you write first. This is a combination.
The formula
The number of ways to choose r items from n items:
With order (permutation): P(n,r) = n! / (n−r)!
Without order (combination): C(n,r) = n! / (r!(n−r)!)
The difference between the formulas is the extra r! in the denominator of the combination. That r! accounts for the r! arrangements of the selected items that are all considered the same outcome (since order does not matter).
Why the formulas have this relationship
If you select 3 people from 10 in order (permutation), you get P(10,3) = 10 × 9 × 8 = 720 outcomes. But for each set of 3 people, there are 3! = 6 ways to arrange them. If order does not matter, all 6 arrangements of the same 3 people count as one outcome. So the number of combinations is 720 ÷ 6 = 120 = C(10,3).
This relationship — C(n,r) = P(n,r) / r! — is not a definition. It is a consequence of what the two formulas are counting.
Worked example: choosing a PIN
A 4-digit PIN uses digits 0–9. How many PINs consist of 4 different digits?
Order matters (1234 is a different PIN from 4321), so this is a permutation.
P(10,4) = 10! / 6! = 10 × 9 × 8 × 7 = 5,040.
Worked example: choosing a study group
From a class of 25 students, a study group of 4 is formed. How many groups are possible?
Order does not matter (the group {Alice, Bob, Carol, David} is the same group regardless of who was listed first), so this is a combination.
C(25,4) = 25! / (4! × 21!) = (25 × 24 × 23 × 22) / (4 × 3 × 2 × 1) = 12,650.
The cases that look like one but are the other
A common trick problem: "How many ways can 5 people sit in 5 chairs in a row?" This looks like a combination (we are choosing from the same people), but the chairs have positions. Person A in chair 1 is different from Person A in chair 3. Order matters. P(5,5) = 5! = 120.
Another: "How many ways can you choose a 2-card hand from a standard 52-card deck?" A hand is unordered — you hold both cards simultaneously, and {Ace, King} is the same hand as {King, Ace}. Combination. C(52,2) = 1,326.
Frequently Asked Questions
What is n! (n factorial)?
n! is the product of all positive integers from 1 to n. So 5! = 5 × 4 × 3 × 2 × 1 = 120. By convention, 0! = 1. Factorials grow very quickly — 20! has 19 digits.
Can I have a permutation or combination with repetition?
Yes — these are permutations with repetition (n^r outcomes) and combinations with repetition (C(n+r−1, r) outcomes). They are less commonly tested but appear in probability and combinatorics courses.