Probability and Permutation-Combination: Complete Guide for Competitive Exams

Probability and permutation-combination (P&C) is the one topic in competitive exam mathematics where candidates either score full marks or lose every question. There is rarely a middle ground — and the reason is simple. Most candidates either understand the underlying counting logic or they do not. Those who rely purely on formula memorization without understanding when to apply each formula consistently make wrong choices under exam pressure.

This guide fixes that. Every formula here comes with the trigger — the exact condition in the question that tells you which formula to use. Every worked example shows not just the calculation but the decision-making process: why this formula and not that one.

In SSC CGL Tier 2, P&C and probability contribute 4–6 questions. In CAT, modern math accounts for 2–3 questions. In IBPS PO Mains, probability appears in 2–3 questions.

Part 1: The Counting Foundation — Before Any Formula

Every P&C problem is a counting problem. Before applying any formula, answer two questions:

Question 1 — Does order matter?

• If YES — use Permutation (arrangement)
• If NO — use Combination (selection)

Question 2 — Are repetitions allowed?

• If YES — use nʳ formula
• If NO — use standard P&C formulas

The Fundamental Counting Principle

If Task 1 can be done in m ways and Task 2 can be done in n ways, then both tasks together can be done in m × n ways.

Worked Example 1:
A person has 4 shirts and 3 trousers. How many outfits can he make?

Outfits = 4 × 3 = 12

Addition vs. Multiplication Rule

Multiply when tasks are done together (AND condition).
Add when tasks are alternatives (OR condition).

Worked Example 2:
A student can travel from A to B by 3 buses or 2 trains. How many ways?

Ways = 3 + 2 = 5 (OR condition)

Worked Example 3:
A to B: 3 buses. B to C: 4 trains. Total ways?

Ways = 3 × 4 = 12 (AND condition)

Part 2: Permutations — Arrangements Where Order Matters

The Core Formula

ⁿPᵣ = n! / (n - r)!

Factorial values to memorize:

• 0! = 1
• 1! = 1
• 2! = 2
• 3! = 6
• 4! = 24
• 5! = 120
• 6! = 720
• 7! = 5040

Worked Example 4 — Basic Permutation

In how many ways can 4 students be arranged from a group of 7?

⁷P₄ = 7 × 6 × 5 × 4 = 840

Shortcut: ⁿPᵣ = n × (n-1) × (n-2) ... (r terms)

Worked Example 5 — All Items Arranged

In how many ways can 5 people sit in 5 chairs?

⁵P₅ = 5! = 120

Permutation with Restrictions

Rule — Fixed Position: Fix the restricted item first, then arrange the rest.

Worked Example 6:
6 people in a row, 2 specific people must sit at the ends.

Step 1: Fix 2 people at ends = 2! = 2 ways
Step 2: Arrange remaining 4 = 4! = 24 ways
Total = 2 × 24 = 48

Rule — Items That Must Stay Together: Treat them as one unit.

Worked Example 7:
5 people in a row, 2 specific people must always sit together.

Step 1: Treat 2 as 1 unit → arrange 4 units = 4! = 24 ways
Step 2: 2 people arrange among themselves = 2! = 2 ways
Total = 24 × 2 = 48

Rule — Items That Must Never Be Together: Total − (together arrangements)

Worked Example 8:
5 people in a row, 2 specific people must never sit together.

Total = 5! = 120
Together = 4! × 2! = 48
Answer = 120 − 48 = 72

Part 3: Combinations — Selections Where Order Does Not Matter

The Core Formula

ⁿCᵣ = n! / (r! × (n-r)!)

Key properties:

• ⁿC₀ = 1
• ⁿC₁ = n
• ⁿCₙ = 1
• ⁿCᵣ = ⁿC₍ₙ₋ᵣ₎ — saves calculation time
• ⁿCᵣ + ⁿC₍ᵣ₊₁₎ = ⁽ⁿ⁺¹⁾C₍ᵣ₊₁₎

Worked Example 9 — Basic Combination

Committee of 3 from 8 people?

⁸C₃ = (8 × 7 × 6) / (3 × 2 × 1) = 336 / 6 = 56

Worked Example 10 — Combination with Condition

Committee of 5 from 6 men and 4 women. At least 2 women.

Exactly 2 women: ⁴C₂ × ⁶C₃ = 6 × 20 = 120
Exactly 3 women: ⁴C₃ × ⁶C₂ = 4 × 15 = 60
Exactly 4 women: ⁴C₄ × ⁶C₁ = 1 × 6 = 6

Total = 120 + 60 + 6 = 186

Complement method:
= ¹⁰C₅ − ⁶C₅ − (⁴C₁ × ⁶C₄)
= 252 − 6 − 60 = 186 ✓

Worked Example 11 — Handshake Problem

10 people, every person shakes hands with every other once.

= ¹⁰C₂ = (10 × 9) / 2 = 45

Diagonal Formula Shortcut

Diagonals in n-sided polygon = n(n-3) / 2

Example: Hexagon (n=6): 6 × 3 / 2 = 9

Part 4: Special Arrangements

Circular Permutations

Circular arrangements of n distinct items = (n-1)!

Worked Example 12:
6 people around a circular table?

= (6-1)! = 5! = 120

Necklace/Garland:
= (n-1)! / 2

Worked Example 13:
6 beads in a necklace?

= 5! / 2 = 120 / 2 = 60

Arrangements with Identical Items

Formula: n! / (p! × q! × r! ...)

Worked Example 14:
Arrangements of MISSISSIPPI?

= 11! / (1! × 4! × 4! × 2!) = 39916800 / 1152 = 34650

Arrangements with Repetition Allowed

Total = nʳ (n items, r positions)

Worked Example 15:
3-digit numbers from digits 1–5, repetition allowed?

= 5³ = 125

Part 5: Probability — From Counting to Likelihood

Core Definition

P(Event) = Favorable outcomes / Total outcomes

Basic properties:

• 0 ≤ P(A) ≤ 1 always
• P(certain event) = 1
• P(impossible event) = 0
• P(not A) = 1 − P(A)

Sample Spaces to Memorize

Worked Example 16 — Basic Probability

Die rolled. Probability of number greater than 4?

Favorable = {5, 6} = 2
P = 2/6 = 1/3

Worked Example 17 — Cards

Card drawn. Probability it is a king or a heart?

P(King) = 4/52
P(Heart) = 13/52
P(King AND Heart) = 1/52

P(King OR Heart) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13

Part 6: Probability Rules

Addition Rule

P(A or B) = P(A) + P(B) − P(A and B)

Mutually exclusive events:
P(A or B) = P(A) + P(B)

Worked Example 18:
Two dice rolled. P(sum = 7 or 11)?

Sum 7: 6 outcomes
Sum 11: 2 outcomes
P = (6 + 2) / 36 = 2/9

Multiplication Rule

Independent events: P(A and B) = P(A) × P(B)

Dependent events: P(A and B) = P(A) × P(B|A)

Worked Example 19:
Two cards drawn without replacement. P(both aces)?

P(1st ace) = 4/52
P(2nd ace | 1st ace) = 3/51

P(both) = 4/52 × 3/51 = 1/221

Worked Example 20:
Bag: 5 red, 3 blue. Two draws with replacement. P(both red)?

= 5/8 × 5/8 = 25/64

Part 7: Conditional Probability

P(B|A) = P(A and B) / P(A)

Worked Example 21:
Bag: 4 red, 3 blue, 2 green. One ball drawn — not green. P(red)?

P(red | not green) = (4/9) / (7/9) = 4/7

Part 8: Exam-Specific Patterns

SSC CGL Pattern

Most common types:

• Arrangements with conditions
• Committees with gender constraints
• Probability from standard deck of cards
• Probability with dice

Speed tip: Always check if ⁿCᵣ = ⁿC₍ₙ₋ᵣ₎ applies. ⁸C₆ = ⁸C₂ = 28 — far faster.

CAT Pattern

Advanced types:

• Distribution problems
• Derangements (no item in its original position)

Derangement formula:
D(n) = n! × (1 − 1/1! + 1/2! − 1/3! + ... + (-1)ⁿ/n!)

D(3) = 2, D(4) = 9

IBPS PO Pattern

• Bags with coloured balls (with/without replacement)
• Standard card deck problems
• Two events combined (addition rule)

Time budget: 60–90 sec per question in Prelims, 90–120 sec in Mains.

Quick Reference — All Formulas

3-Week Practice Plan