Precalculus — Chapter 12: Counting & Probability

Combinations & Permutations

Factorial notation, the Fundamental Counting Principle, permutations (order matters), and combinations (order doesn't) — with step-by-step worked examples.

Factorial Notation

n! = n × (n−1) × (n−2) × ··· × 2 × 1

Read as "n factorial". Special case: 0! = 1

0!

1

1!

1

2!

2

3!

6

4!

24

5!

120

6!

720

7!

5,040

8!

40,320

9!

362,880

10!

3,628,800

Fundamental Counting Principle

If one event can occur in m ways and a second independent event can occur in n ways, the two events can occur together in m × n ways.

Total outcomes = n₁ × n₂ × n₃ × ···

Worked Example 1

A license plate has 3 letters followed by 4 digits. How many plates are possible?

Letters: 26 choices each × 3 positions = 26 × 26 × 26

Digits: 10 choices each × 4 positions = 10 × 10 × 10 × 10

Total = 26³ × 10⁴ = 17,576 × 10,000

= 175,760,000 possible plates

Permutations — Order Matters

A permutation is an ordered arrangement. P(n,r) = nPr counts the number of ways to arrange r items chosen from n distinct items.

P(n, r) = n! / (n − r)!

n = total items, r = items selected, order matters

Worked Example 2

How many ways can 8 runners finish 1st, 2nd, and 3rd?

P(8, 3) = 8! / (8−3)! = 8! / 5!

= (8 × 7 × 6 × 5!) / 5!

= 8 × 7 × 6

= 336 arrangements

Worked Example 3

How many 4-letter codes from the word MATH (no repeats)?

n = 4 letters, r = 4 positions

P(4, 4) = 4! / (4−4)! = 4! / 0!

= 24 / 1

= 24 arrangements

Permutations with Repeated Elements

When some items are identical, divide by the factorial of each repeated group:

n! / (n₁! × n₂! × ··· × nₖ!)

Example: Arrangements of MISSISSIPPI (11 letters: 1 M, 4 I, 4 S, 2 P) = 11! / (1! × 4! × 4! × 2!) = 34,650

Combinations — Order Doesn't Matter

A combination counts the number of ways to choose r items from n distinct items without regard to order.

C(n, r) = n! / (r! × (n − r)!)

Also written C(n,r), ⁿCᵣ, or {n choose r}

Worked Example 4

How many ways to choose 3 students from a class of 20 for a committee?

C(20, 3) = 20! / (3! × 17!)

= (20 × 19 × 18) / (3 × 2 × 1)

= 6,840 / 6

= 1,140 combinations

Worked Example 5

A 5-card hand from a 52-card deck — how many hands?

C(52, 5) = 52! / (5! × 47!)

= (52 × 51 × 50 × 49 × 48) / (5!)

= 311,875,200 / 120

= 2,598,960 hands

C(n,0) = 1

One way to choose nothing

C(n,n) = 1

One way to choose everything

C(n,r) = C(n, n−r)

Symmetry: choosing r is same as excluding n−r

Permutation vs Combination — Decision Guide

ScenarioOrder Matters?Formula
Race finishing positions (1st, 2nd, 3rd)YesP(n, r)
Seating arrangement at a tableYesP(n, r) or n!
Password / PIN digitsYesP(n, r) or nʳ with repetition
Committee selection (no roles)NoC(n, r)
Lottery ticket (pick 6 of 49)NoC(n, r)
Card hand from a deckNoC(n, r)
Pizza toppings (choose 3 of 10)NoC(n, r)
Arrange letters in a wordYesn! or n!/repeats

Multi-Step Counting Problem

Worked Example 6

A club has 10 members. How many ways can they elect a president, VP, and a 3-person committee (with no person holding two roles)?

Step 1 — Elect president and VP (order matters)

P(10, 2) = 10 × 9 = 90

Step 2 — Choose 3 from remaining 8 (order doesn't matter)

C(8, 3) = 8!/(3!×5!) = 56

Step 3 — Multiply (Fundamental Counting Principle)

90 × 56 = 5,040

= 5,040 ways

Formula Quick Reference

Factorialn! = n×(n−1)×···×1, 0!=1
Fundamental Countingn₁ × n₂ × ··· × nₖ
Permutation P(n,r)n! / (n−r)!
Combination C(n,r)n! / [r!(n−r)!]
Permutations w/ repeatsn! / (n₁! n₂! ··· nₖ!)
RelationC(n,r) = P(n,r) / r!

Exam Strategy

Ask: does swapping matter?

If swapping two items gives a different valid outcome, use P(n,r). If swapping gives the same result, use C(n,r). "Committee of 3" → same; "1st, 2nd, 3rd place" → different.

Cancel factorials in the formula

In P(n,r) = n!/(n−r)!, the (n−r)! cancels with the bottom of n!, leaving just n×(n−1)×···×(n−r+1). Never compute the full factorial if you can cancel.

Break multi-stage problems apart

Identify each independent choice, compute P or C for each stage, then multiply them all together using the Fundamental Counting Principle.

Frequently Asked Questions

What is the difference between a permutation and a combination?

A permutation is an arrangement where order matters (1st, 2nd, 3rd are different outcomes). A combination is a selection where order does NOT matter (choosing 3 people for a committee — the order they're chosen is irrelevant). As a rule: if you can swap two elements and get a different outcome, it's a permutation.

When do I use nCr vs nPr on my calculator?

Use nPr (permutations) when arrangement matters — passwords, race finishing positions, seating arrangements. Use nCr (combinations) when only the group matters — lottery tickets, committee selections, card hands. nCr = nPr ÷ r! because combinations discard the r! orderings of each group.

What is 0! (zero factorial)?

0! = 1 by definition. This is required for the formulas to work correctly — for example, C(n,0) = n!/(0!·n!) = 1, meaning there is exactly one way to choose nothing from a set. The result 0! = 1 is also consistent with the recursive definition n! = n × (n−1)!.

Related Topics

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