The one question that decides it
When you count ways to pick k things from n, ask: if I swap two of the chosen things, do I get a different outcome?
If yes, order is part of the outcome, and you are counting arrangements, called permutations. If no, only the group matters, and you are counting selections, called combinations. Nearly every formula choice follows from that one question.
Permutations: order is part of the outcome
Take 8 runners competing for gold, silver and bronze. Alice-Bob-Carol is a different result from Carol-Bob-Alice, so each ordered podium counts separately.
- 8 choices for gold
- 7 for silver
- 6 for bronze
8 × 7 × 6 = 336 podiums.
The general formula for ordered selections of k from n is n × (n − 1) × … × (n − k + 1), which equals n! ÷ (n − k)!. For this example, 8! ÷ 5! = 40,320 ÷ 120 = 336, which agrees.
Other situations where order matters:
- A PIN or lock setting such as 4-7-1, as with combination locks.
- A turn order in a board game with 5 players: 5! = 120 possible orders.
- A captain and a vice-captain from 10 players: 10 × 9 = 90, since the two roles differ.
Combinations: only the group matters
Now suppose the three runners are simply picked as a relay group, with no ranking. Alice, Bob and Carol is the same group however you list them.
Each group of three can be listed in 3! = 6 orders, and the 336 ordered podiums count each group six times. So
336 ÷ 6 = 56 groups.
The general formula is the permutation count divided by k!:
C(n, k) = n! ÷ (k! × (n − k)!).
Check: C(8, 3) = 8! ÷ (3! × 5!) = 40,320 ÷ (6 × 120) = 40,320 ÷ 720 = 56.
More situations where order does not matter:
- Choosing 2 players out of 10 to form a pair: C(10, 2) = 10 × 9 ÷ 2 = 45, compared with 90 ordered captain-and-vice pairs.
- Picking 3 of 5 equipment items to carry: 5 × 4 × 3 ÷ 6 = 10.
- Dealing a five-card hand: 311,875,200 ordered deals divided by 5! = 120 gives 2,598,960 hands.
Seeing the link: divide by k!
The relationship between the two is simple: permutations = combinations × k!. A combination is a permutation with the order ignored, so you divide out the k! ways to order the chosen group. The same logic shows why C(n, k) = C(n, n − k): choosing k to take is the same decision as choosing n − k to leave behind. Small values are collected in Pascals triangle.
| n = 5, k = 3 | Count |
|---|---|
| Ordered (permutations) | 5 × 4 × 3 = 60 |
| Unordered (combinations) | 60 ÷ 6 = 10 |
Spotting which one a problem wants
Words that usually suggest order matters: arrange, rank, line up, schedule, code, password, first/second/third, assign distinct roles.
Words that usually suggest it does not: choose, select, committee, group, hand, team of equals, subset.
Wording can mislead, so test with the swap question. “Pick 2 people from 5 to shake hands” is unordered, but “pick one person to shake hands with another” may or may not be depending on how the problem treats the two roles. When in doubt, imagine two outcomes that differ only by order and ask whether you would call them the same.
Common mistakes
Using combinations for a lock. A lock with settings 2-5-8 and 8-5-2 opens for exactly one of them, so these are different codes. Using C(10, 3) = 120 would undercount; 10 × 9 × 8 = 720 (distinct digits) or 10³ = 1,000 (repeats allowed) is right.
Using permutations for a hand. A poker hand does not care about arrival order. Counting ordered deals overstates the number of hands by a factor of 120.
Forgetting repetition. Both formulas above assume each item can be used once. If items can repeat, permutations become n^k, and unordered selections with repeats need a different tool, stars and bars.
Forgetting identical items. If some items are identical, such as the letters in “LEVEL”, orderings that swap identical items are not distinct. You divide by the factorial of each repeat count: 5! ÷ (2! × 2!) = 120 ÷ 4 = 30.
Why it matters for probability
When outcomes are equally likely, probability is favourable divided by total. If you count the numerator as ordered and the denominator as unordered, or the other way around, the answer is wrong. For example, the chance that a three-person group drawn from 8 contains two particular people is the number of groups containing them, 6 (the third member can be any of the other 6), divided by 56, giving 6/56 = 3/28, about 10.7 percent. Counting ordered podiums throughout would give the same ratio, but mixing the two would not.
Both counts are special cases of the multiplication principle in the wider overview of counting principles, and the vocabulary also turns up in discussions of combinations and permutations in game settings. The aim is to count accurately, which is separate from predicting any single result: shuffles and rolls are independent, and games of chance with a built-in house advantage can cost a player money over time. Anyone finding gambling hard to control should consider contacting a local support service.