Combinations & Permutations
What is the Combinations & Permutations?
Combinations and permutations both count the number of ways to select items from a group, but they differ in one crucial way: combinations don't care about order, permutations do. Choosing 3 pizza toppings from a menu of 10 is a combination โ the order you pick them in doesn't matter, since "pepperoni, mushroom, olive" is the same pizza as "olive, pepperoni, mushroom." Arranging 3 runners on a podium from a field of 10 is a permutation โ first, second, and third place are all different outcomes, so who finishes where genuinely matters.
Because permutations count every possible ordering as distinct, P(n,r) is always at least as large as C(n,r) for the same n and r โ in fact, P(n,r) = C(n,r) ร r!, since each combination can be arranged in r! different orders. Both formulas rely on the factorial function (n!), which grows extraordinarily fast โ 10! is already over 3.6 million, which is part of why even modest counting problems can have surprisingly large answers.
These two formulas are the foundation of combinatorics and probability more broadly: almost any "how many ways can this happen" question, from lottery odds to password strength to the number of possible card hands, reduces to one of these two formulas, or a combination of both applied to different parts of the same problem.
What Each Variable Means
Units
| Quantity | Symbol | Unit |
|---|---|---|
| Combinations | C(n,r) | n! / (r!(n-r)!) |
| Permutations | P(n,r) | n! / (n-r)! |
When to Use It
- Combinations โ when selecting a group where the order doesn't matter (a committee, a hand of cards, toppings)
- Permutations โ when the order or arrangement matters (a ranking, a race finish, a PIN code)
- Probability problems that require counting the total number of possible outcomes
Step-by-Step Example
Problem: How many ways can you choose 3 people from a group of 5 (a) as an unordered group, and (b) as an ordered arrangement?
Use C(n,r) = n! / (r!(n-r)!).
C(5,3) = 5! / (3! ร 2!) = 120 / (6ร2) = 10Use P(n,r) = n! / (n-r)!.
P(5,3) = 5! / 2! = 120 / 2 = 60Interactive Calculator
Common Mistakes
Mistake: Using the permutation formula when order doesn't actually matter (or vice versa).
Fix: Always ask first: does rearranging the same items count as a different outcome? If yes, use permutations; if no, use combinations.
Mistake: Forgetting that 0! = 1, not 0.
Fix: By definition, 0! = 1 โ this matters when r = n, since (nโr)! becomes 0! in the formula.
Practice Questions
How many ways can you choose 2 books from a shelf of 6 (order doesn't matter)?
How many ways can 4 runners finish in 1st, 2nd, and 3rd place?
Hint: Order matters here, since the placements are distinct.
Frequently Asked Questions
How are combinations and permutations related?
P(n,r) = C(n,r) ร r! โ every combination of r items can be arranged in r! different orders, so permutations always count at least as many outcomes as combinations.
What if r is 0?
Both C(n,0) and P(n,0) equal 1 โ there's exactly one way to choose nothing at all.