What are Permutation, Combination & Probability? These are counting and chance calculation topics. Permutation = arrangements (order matters), Combination = selections (order doesn't matter), Probability = chance of event happening. All three are interconnected through counting principles.

A
B
C
D
4! = 24 ways

Pro Tip – Remember the Key Difference!

Permutation = ARRANGEMENT (Order matters: ABC ≠ CBA). Combination = SELECTION (Order doesn't matter: ABC = CBA). Visit SKY Practice for 500+ P&C questions with decision trees.

1. Fundamental Counting Principles

Basic Rules: These principles form the foundation of all counting problems.

Core Counting Rules

Understanding Counting Principles

Before diving into formulas, understand these basic rules that help solve most counting problems logically.

Multiplication Principle

If one operation can be done in 'm' ways and another in 'n' ways, then both can be done in:

m × n ways
Example: 3 shirts × 4 pants = ?
3 × 4 = 12 different outfits

Addition Principle

If one operation can be done in 'm' ways OR another in 'n' ways (mutually exclusive), then total ways:

m + n ways
Example: Travel by 3 trains OR 4 buses = ?
3 + 4 = 7 ways to travel

Factorial Notation

n! = n × (n-1) × (n-2) × ... × 3 × 2 × 1

Special values:
0! = 1
1! = 1
5! = 120
6! = 720
10! = 3,628,800

SSC Shortcut: Quick Factorial Values

Must remember: 5! = 120, 6! = 720, 7! = 5040, 8! = 40320

For division: 8!/6! = 8×7 = 56 (cancel 6!)

Example: 10!/8! = 10×9 = 90

Combination: ⁷C₃ = 7!/(3!4!) = (7×6×5)/(3×2×1) = 35

Solved Example: Counting Principle

Q: How many 3-digit numbers can be formed from digits 1,2,3,4,5 without repetition?
Step 1: Identify positions
3-digit number: Hundreds, Tens, Units places

Step 2: Apply multiplication principle
Hundreds place: 5 choices (1,2,3,4,5)
Tens place: 4 choices (one digit used)
Units place: 3 choices (two digits used)

Step 3: Multiply
Total numbers = 5 × 4 × 3 = 60

Alternative using permutation formula:
⁵P₃ = 5!/(5-3)! = 5!/2! = (5×4×3×2×1)/(2×1) = 5×4×3 = 60

List few numbers to verify:
123, 124, 125, 132, 134, 135, 142, 143, 145, 152...
Each digit appears in each position equally

Final Answer: 60 three-digit numbers

2. Permutation (nPr) - Arrangements

What is Permutation? Arrangement of objects where ORDER MATTERS. ABC is different from CBA.

Permutation Formulas

Understanding Permutation

Use permutation when you need to arrange/order things. Common keywords: arrange, form numbers, seating arrangements, ranking.

Type Formula Notation Example Simple Permutation ⁿPᵣ = n!/(n-r)! Arranging r out of n ⁵P₃ = 5!/(5-3)! = 60 All n items ⁿPₙ = n! All items arranged 5! = 120 arrangements With Repetition nʳ r positions, n choices each 3² = 9 two-digit numbers from 1,2,3 Circular Arrangement (n-1)! Around circle (fixed relative) 5 people around table = 4! = 24 With Identical Items n!/(p!q!r!...) p,q,r identical items ARRANGE: 7!/(2!2!) = 1260
A
B
C
C
B
A

ABC ≠ CBA → Order matters → Permutation

SSC Shortcut: Permutation Values to Remember

⁵P₂ = 20, ⁵P₃ = 60, ⁶P₂ = 30, ⁶P₃ = 120

⁷P₂ = 42, ⁷P₃ = 210, ⁸P₂ = 56, ⁸P₃ = 336

Circular: (n-1)! → 5 people = 4! = 24, 6 people = 5! = 120

Solved Example: Complex Permutation

Q: In how many ways can the letters of the word "BANANA" be arranged?
Step 1: Analyze the word
BANANA has 6 letters:
B: 1 time, A: 3 times, N: 2 times

Step 2: Apply formula for identical items
When items are identical, divide by factorial of counts:
Total arrangements = 6!/(3!2!1!)

Step 3: Calculate
6! = 720
3! = 6, 2! = 2, 1! = 1
Arrangements = 720/(6×2×1) = 720/12 = 60

Step 4: Verify with smaller example
Word "AA" (2 letters, both A):
Formula: 2!/2! = 1 (only AA)
Word "AB" (2 distinct): 2! = 2 (AB, BA)

Alternative: List to understand
If all letters were distinct: 6! = 720 ways
But 3 A's are identical: divide by 3! = 6
And 2 N's are identical: divide by 2! = 2
So 720/(6×2) = 720/12 = 60

Final Answer: 60 different arrangements

3. Combination (nCr) - Selections

What is Combination? Selection of objects where ORDER DOESN'T MATTER. Choosing {A,B,C} is same as {C,B,A}.

Combination Formulas

Understanding Combination

Use combination when you need to select/choose things. Common keywords: select, choose, form committee, pick team.

Type Formula Notation Example Simple Combination ⁿCᵣ = n!/(r!(n-r)!) Choose r from n ⁵C₃ = 5!/(3!2!) = 10 All items ⁿCₙ = 1 Choose all Only 1 way to choose all None ⁿC₀ = 1 Choose none Only 1 way to choose none Complementary ⁿCᵣ = ⁿCₙ₋ᵣ Choosing = Not choosing ⁵C₂ = ⁵C₃ = 10 With Restrictions Case analysis Include/exclude specific Choose 3 from 5 with A included

SSC Shortcut: Combination Values to Remember

⁵C₂ = 10, ⁵C₃ = 10, ⁶C₂ = 15, ⁶C₃ = 20

⁷C₂ = 21, ⁷C₃ = 35, ⁸C₂ = 28, ⁸C₃ = 56

Pascal's Triangle property: ⁿCᵣ + ⁿCᵣ₊₁ = ⁿ⁺¹Cᵣ₊₁

Sum of row: ⁿC₀ + ⁿC₁ + ... + ⁿCₙ = 2ⁿ

Solved Example: Complex Combination

Q: From 6 men and 4 women, a committee of 5 is to be formed. In how many ways can it be done if the committee must have at least 3 women?
Step 1: Understand "at least 3 women"
This means: 3 women OR 4 women (can't have 5 women as only 4 available)
Cases: 3W+2M OR 4W+1M

Step 2: Case 1: 3 women, 2 men
Choose 3 women from 4: ⁴C₃ = 4
Choose 2 men from 6: ⁶C₂ = 15
Ways for Case 1 = 4 × 15 = 60

Step 3: Case 2: 4 women, 1 man
Choose 4 women from 4: ⁴C₄ = 1
Choose 1 man from 6: ⁶C₁ = 6
Ways for Case 2 = 1 × 6 = 6

Step 4: Add both cases
Total ways = 60 + 6 = 66

Step 5: Verify with complementary method
Total ways without restriction: Choose 5 from 10 = ¹⁰C₅ = 252
Ways with 0 women (5 men): ⁶C₅ × ⁴C₀ = 6 × 1 = 6
Ways with 1 woman (4 men): ⁴C₁ × ⁶C₄ = 4 × 15 = 60
Ways with 2 women (3 men): ⁴C₂ × ⁶C₃ = 6 × 20 = 120
Sum of invalid cases = 6 + 60 + 120 = 186
Valid ways = 252 - 186 = 66 ✓

Final Answer: 66 different committees

4. Probability Basics

What is Probability? Measure of likelihood of an event occurring, between 0 (impossible) and 1 (certain).

Probability Formulas

Understanding Probability

Probability = Favorable outcomes / Total possible outcomes. All outcomes must be equally likely.

Sample Space

Total outcomes = 6

1 2 3 4 5 6
Even: 2,4,6

P(Even) = 3/6 = 1/2

Basic Probability

P(E) = n(E)/n(S)

Where:
P(E) = Probability of event E
n(E) = Number of favorable outcomes
n(S) = Total possible outcomes

Die: P(Prime) = ?
Prime numbers: 2,3,5 (3 outcomes)
Total outcomes: 6
P(Prime) = 3/6 = 1/2

Complement Rule

P(E') = 1 - P(E)

Where E' = event not happening

P(Rain) = 0.3, P(No Rain) = ?
P(No Rain) = 1 - 0.3 = 0.7

Range of Probability

  • 0 ≤ P(E) ≤ 1
  • P(Impossible) = 0
  • P(Certain) = 1
  • Sum of all probabilities = 1
  • Odds in favor = P(E)/P(E')
  • Odds against = P(E')/P(E)

SSC Shortcut: Common Probability Values

Coin: P(Head) = 1/2, P(at least 1 head in 2 tosses) = 3/4

Die: P(Even) = 1/2, P(Prime) = 1/2, P(≥4) = 1/2

Cards: P(Heart) = 13/52 = 1/4, P(Ace) = 4/52 = 1/13

Complement: P(at least 1) = 1 - P(none)

Solved Example: Probability Problem

Q: Two dice are thrown simultaneously. What is the probability that the sum is 9?
Step 1: Find total outcomes
Each die has 6 faces
Total outcomes = 6 × 6 = 36

Step 2: Find favorable outcomes (sum = 9)
Pairs: (3,6), (4,5), (5,4), (6,3)
Count = 4 favorable outcomes

Step 3: Calculate probability
P(Sum=9) = 4/36 = 1/9

Step 4: Verify all sums
Minimum sum = 2 (1,1)
Maximum sum = 12 (6,6)
Most likely sum = 7 (6 ways)

Alternative: Sample space table
Create 6×6 grid: Rows = Die1, Columns = Die2
Mark cells where sum = 9: 4 cells out of 36

Final Answer: Probability = 1/9

5. Conditional Probability

What is Conditional Probability? Probability of event A given that event B has already occurred.

Conditional Probability Formulas

Understanding "Given That"

P(A|B) means probability of A happening, given that B has already happened. The sample space reduces to only those outcomes where B occurred.

Start
B occurs
P(B)
A occurs
P(A|B)
B' occurs
P(B')
A occurs
P(A|B')

Conditional Formula

P(A|B) = P(A∩B)/P(B)

Where P(B) > 0

P(Rain)=0.3, P(Umbrella|Rain)=0.8
P(Rain∩Umbrella) = 0.3×0.8 = 0.24

Multiplication Rule

P(A∩B) = P(A) × P(B|A)
= P(B) × P(A|B)

For independent events:
P(A∩B) = P(A) × P(B)

Independent Events

  • A and B independent if P(A∩B) = P(A)×P(B)
  • Equivalent: P(A|B) = P(A)
  • Equivalent: P(B|A) = P(B)
  • Coin tosses are independent
  • Cards without replacement: dependent

SSC Shortcut: Cards Probability

With replacement: Independent events

Without replacement: Dependent events

Example: 2 cards from deck without replacement:

P(First Ace) = 4/52 = 1/13

P(Second Ace | First Ace) = 3/51 = 1/17

P(Both Aces) = (4/52)×(3/51) = 1/221

Solved Example: Conditional Probability

Q: In a class, 60% students like Math, 40% like Science, and 30% like both. If a student likes Science, what is the probability they also like Math?
Step 1: Define events
Let M = likes Math, S = likes Science
Given: P(M) = 0.60, P(S) = 0.40, P(M∩S) = 0.30

Step 2: Apply conditional probability formula
We need P(M|S) = Probability likes Math given likes Science
P(M|S) = P(M∩S)/P(S)

Step 3: Substitute values
P(M|S) = 0.30/0.40 = 3/4 = 0.75

Step 4: Interpret result
75% of Science students also like Math

Step 5: Verify with numbers
Assume 100 students total:
Like Math: 60 students
Like Science: 40 students
Like both: 30 students
Given student likes Science (40 students), how many like Math? 30
So probability = 30/40 = 0.75 ✓

Final Answer: Probability = 0.75 or 75%

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Includes all types: counting principles, permutations, combinations, probability, conditional probability

Frequently Asked Questions

Q1: How many P&C Probability questions in SSC CGL?

Answer: Typically 4-6 questions in Tier I and 8-10 questions in Tier II. These are moderate difficulty questions focusing on application of formulas rather than complex theory.

Q2: What's the easiest way to remember nPr vs nCr?

Answer: P for Permutation = Positions (order matters). C for Combination = Choose (order doesn't matter). Permutation has more arrangements: ⁿPᵣ = ⁿCᵣ × r!

Q3: When to use permutation vs combination in word problems?

Answer: Keywords: "Arrange, order, sequence" → Permutation. "Select, choose, form committee, pick team" → Combination. If swapping creates new arrangement → Permutation.

Q4: How to handle "at least" or "at most" probability questions?

Answer: Use complement rule: P(at least 1) = 1 - P(none). For "at most", add probabilities from 0 up to the limit. Often easier than adding many cases.

Q5: What's the difference between independent and mutually exclusive events?

Answer: Independent: Occurrence of one doesn't affect other (P(A∩B)=P(A)×P(B)). Mutually exclusive: Cannot occur together (P(A∩B)=0). Independent events can occur together, mutually exclusive cannot.

Q6: How to solve circular arrangement problems quickly?

Answer: Fix one person's position to break rotational symmetry. Then arrange remaining (n-1)! ways. For necklace/flip symmetry: (n-1)!/2.

Final Exam Strategy for P&C Probability

Time Allocation: Basic problems: 40-60 seconds, Complex problems: 90-120 seconds.

Priority Order: 1) Basic counting, 2) Probability, 3) Combinations, 4) Permutations, 5) Conditional probability.

Accuracy Check: Verify if answer should be ≤ total possible. For probability, check 0≤answer≤1. Use small numbers to test logic.

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