Permutations and Combinations

Distinguish ordered arrangements from unordered selections and count each correctly.

1. The idea

A permutation counts ordered arrangements: choosing and arranging $k$ items from $n$ distinct items gives $P (n, k) = \frac{n !}{( n - k )!}$ possibilities, since order matters.

A combination counts unordered selections: choosing $k$ items from $n$ without regard to order gives $C (n, k) = (k n) = \frac{n !}{k ! ( n - k )!}$ . Every combination corresponds to $k!$ permutations, which is why dividing by $k!$ converts one count into the other.

The single most important habit is asking "does order matter here?" before choosing a formula. Seating arrangements, rankings, and passwords are ordered; committees, hands of cards, and unordered subsets are not.

2. Why it matters

Nearly every combinatorics and discrete-probability question starts with correctly identifying whether you're counting ordered or unordered objects — getting this wrong invalidates the entire rest of the solution.

3. Core mathematics

Permutations

P (n, k) = \frac{n !}{( n - k )!}

Combinations

(k n) = \frac{n !}{k ! ( n - k )!}

Relationship

P (n, k) = k! (k n)

Symmetry identity

(k n) = (n - k n)

4. How to recognize it

Think of this when…

The question mentions arranging, ranking, or ordering a subset of items — use permutations.
The question mentions choosing a committee, subset, or hand — use combinations.
Passwords, license plates, and seatings are ordered; teams and lottery draws are unordered.
You are asked for the number of ways to pick $k$ from $n$ without further arrangement.

5. Worked example

Worked example

From 10 distinct books, how many ways can you (a) arrange 3 of them on a shelf, and (b) choose 3 of them to give away (order doesn't matter)?

1
For (a), order matters (shelf position), so use permutations.
$P (10, 3) = \frac{10 !}{7 !} = 10 \times 9 \times 8 = 720$
2
For (b), order doesn't matter, so use combinations.
$(3 10) = \frac{10 !}{3 ! 7 !} = \frac{10 \times 9 \times 8}{6} = 120$
3
Check consistency: each combination corresponds to $3! = 6$ orderings, and indeed $120 \times 6 = 720$ .
Answer: (a) 720 arrangements; (b) 120 selections.

6. Common mistakes

Using $(k n)$ when the problem actually requires an arrangement (order matters).
Using $P (n, k)$ when items chosen are functionally identical or unordered, overcounting by a factor of $k!$ .
Forgetting $0! = 1$ , which matters when $k = 0$ or $k = n$ .
Double-counting when items chosen from distinguishable groups are then treated as interchangeable, or vice versa.

7. Quick check

How many ways can you choose 4 people from 10 to form an unordered committee?

How many ways can 4 runners finish 1st through 4th out of 10 runners (all distinct finishing times)?

8. Practice

46 QuantSolve questions use this concept.

Practice this concept

9. Related methods

Complement Rule

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