One question decides it
Before counting a selection, ask: if I swap two of the chosen items, do I get a different result?
- If yes, order matters, and you are counting permutations.
- If no, order does not matter, and you are counting combinations.
Everything else follows from that. A combination is simply a permutation count with the duplicate orderings divided out.
Permutations: order matters
Seating
Five friends sit in five chairs in a row. The first chair can be filled 5 ways, the second 4 ways, then 3, 2 and 1. The total is 5 × 4 × 3 × 2 × 1 = 5! = 120 seatings. Swapping any two friends gives a different arrangement, so order matters.
Partial arrangements
Now imagine 5 runners and only the gold, silver and bronze places are recorded. There are 5 choices for gold, 4 for silver and 3 for bronze:
5 × 4 × 3 = 60
In notation, P(5, 3) = 5! ÷ 2! = 120 ÷ 2 = 60. The general formula is P(n, k) = n! ÷ (n − k)!.
Passwords and PIN codes follow the same ordered logic, although there you may be allowed to reuse characters, which changes the count. When repeats are allowed, you simply multiply the number of options at each position, as in the multiplication principle: a 4-digit PIN has 10 × 10 × 10 × 10 = 10,000 possibilities.
Combinations: order does not matter
Teams and committees
Pick a team of 3 from the same 5 people, with no roles. The team {Ana, Ben, Cy} is the same team however you list the names. Each team of 3 appears in the 60 ordered results 3! = 6 times, because 3 people can be ordered in 6 ways. So:
60 ÷ 6 = 10 teams
In notation, C(5, 3) = 5! ÷ (3! × 2!) = 120 ÷ 12 = 10. The general formula is C(n, k) = n! ÷ (k! × (n − k)!), which is P(n, k) ÷ k!.
Card hands
A five-card hand from a standard deck is a combination, because the order you receive the cards in does not change what you hold. If order did matter, there would be 52 × 51 × 50 × 49 × 48 = 311,875,200 ordered deals. Each hand can be arranged 5! = 120 ways, so:
311,875,200 ÷ 120 = 2,598,960 different hands
The same reasoning gives the 1,326 possible two-card starting hands: 52 × 51 = 2,652 ordered pairs, divided by 2 = 1,326.
Lottery picks
In a typical lottery, you choose a set of numbers, and what matters is which numbers are drawn, not the order of the draw. A format drawing 6 numbers from 49 gives C(49, 6) = 13,983,816 possible sets. If order mattered, the count would be 720 times larger, since 6! = 720. This is the basis of lottery odds: the odds come from counting the sets and are not changed by which numbers you choose.
A comparison table
| Situation | Order matters? | Count |
|---|---|---|
| 5 friends in 5 chairs | Yes | 5! = 120 |
| Gold, silver, bronze among 5 runners | Yes | 5 × 4 × 3 = 60 |
| Team of 3 from 5 people | No | C(5, 3) = 10 |
| 5-card hand from 52 | No | 2,598,960 |
| 6 numbers from 49 | No | 13,983,816 |
Tricky cases
Real problems can mix both ideas. Say a club has 6 members and wants to choose a president, a treasurer and a secretary, then also pick two other members as general helpers from those who remain. Roles make the first part ordered: 6 × 5 × 4 = 120. The helpers are unranked, so the second part is C(3, 2) = 3. Because the choices happen in stages, you multiply: 120 × 3 = 360.
A helpful habit is to describe the outcome in words. “Who gets which role” is ordered. “Which people are in the group” is unordered.
Common mistakes
- Using a permutation when a combination is needed. This overstates a count by a factor of k!.
- Assuming a “combination lock” is a combination. A lock code is an ordered sequence, so it is really a permutation, often with repeats allowed.
- Forgetting that sampling without replacement shrinks the options. Each pick leaves one fewer item available, as covered when drawing without replacement.
- Dividing by the wrong number. You divide by the number of ways to order the chosen items, k!, not by n!.
Why this matters for odds
Probability questions about hands, draws and teams usually reduce to dividing one count by another. If the numerator and denominator are counted with different assumptions about order, the answer is wrong. Staying consistent is what counts, and either all-ordered or all-unordered works as long as you use it for both.
The counts above describe how many possibilities exist. They do not make any game easier to beat. Games with a built-in house edge lose money over time, and gambling can be harmful.