# Permutations and Combinations — When to Arrange and When to Choose

Source: https://skills101.in/learn/aptitude/permutations-and-combinations — Skills101 free tutorial (Quantitative Aptitude). Updated 2026-11-09.

This chapter is about **counting the ways** something can happen — without listing them all.

## 1. The counting principle

> If one thing can be done in **m** ways and, after it, another in **n** ways, both can be done in **m × n** ways (AND → multiply). If you do **either** one **or** the other, it is **m + n** ways (OR → add).

### Example 1
*A shirt comes in 4 colours and 3 sizes. How many different shirts?* 4 × 3 = **12**.

### Example 2
*You can travel Jaipur → Delhi by 3 trains or 2 buses. How many ways?* 3 + 2 = **5**.

## 2. Factorial

n! = n × (n − 1) × … × 2 × 1. So 5! = 120, 4! = 24, 3! = 6, 0! = **1**.

## 3. Arrange (permutation) vs choose (combination)

![Arranging 3 of 5 people: slots of 5, 4, 3 choices give 60 arrangements. Choosing a team of 3: the 6 orders of the same 3 people are one team, so 60 ÷ 6 = 10](https://skills101.in/learn-img/aptitude/pnc-slots.svg)

| | Order matters? | Formula | Keywords |
|---|---|---|---|
| **Permutation** | Yes | ⁿPᵣ = n! ÷ (n − r)! | arrange, order, rank, password, seating, digits of a number |
| **Combination** | No | ⁿCᵣ = n! ÷ (r! (n − r)!) | select, choose, team, committee, group, handshakes |

> **ⁿCᵣ = ⁿPᵣ ÷ r!** — choosing is arranging, then ignoring the r! orders.

Useful facts: ⁿCᵣ = ⁿCₙ₋ᵣ (choosing 8 of 10 = leaving out 2 of 10), ⁿC₀ = ⁿCₙ = 1.

### Example 3
*How many 3-digit numbers can be made from 1, 2, 3, 4, 5 without repeating a digit?*

Order matters: ⁵P₃ = 5 × 4 × 3 = **60**.

### Example 4
*In how many ways can a committee of 3 be chosen from 8 people?*

⁸C₃ = (8 × 7 × 6) ÷ (3 × 2 × 1) = **56**.

## 4. Arranging letters of a word

n letters, all different → n! ways. If some letters repeat, divide by the factorial of each repeat count.

### Example 5
*How many arrangements of the letters of LEVEL?*

5 letters: L twice, E twice → 5! ÷ (2! × 2!) = 120 ÷ 4 = **30**.

### Example 6 — letters together
*In how many arrangements of the word APPLE do the vowels come together?*

Vowels A, E → tie them as one block. Items: [AE], P, P, L = 4 items with P twice → 4!/2! = 12. Inside the block, A and E swap: × 2 → **24**.

## 5. Circular arrangements

Around a round table, rotating everyone gives the same seating, so fix one person: **(n − 1)!** ways.

### Example 7
*In how many ways can 6 friends sit around a round table?* (6 − 1)! = **120**.

## 6. Committees with conditions

### Example 8
*From 5 men and 4 women, a committee of 3 must have at least 1 woman. How many ways?*

Total − (no women) = ⁹C₃ − ⁵C₃ = 84 − 10 = **74**.

> **"At least one" → total minus "none".** Much faster than adding cases.

### Example 9 — handshakes
*10 people each shake hands once with everyone else. How many handshakes?* ¹⁰C₂ = **45**.

## Common mistakes

- Using P when the order doesn't matter (teams, groups) — or C when it does (passwords, rankings).
- Forgetting to divide for repeated letters.
- Forgetting to arrange the people **inside** a "together" block.

## Quick recap

- AND → multiply, OR → add.
- Order matters → ⁿPᵣ; order doesn't → ⁿCᵣ = ⁿPᵣ / r!.
- Repeated letters: n! ÷ (p! q! …).
- Round table: (n − 1)!.
- "At least one" = total − none.


Practice quiz and free certificate: https://skills101.in/learn/aptitude/permutations-and-combinations