Permutation, Combination & Probability - Complete SSC CGL Guide
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.
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:
Addition Principle
If one operation can be done in 'm' ways OR another in 'n' ways (mutually exclusive), then total ways:
Factorial Notation
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
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.
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
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.
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
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
P(Even) = 3/6 = 1/2
Basic Probability
Where:
P(E) = Probability of event E
n(E) = Number of favorable outcomes
n(S) = Total possible outcomes
Total outcomes: 6
P(Prime) = 3/6 = 1/2
Complement Rule
Where E' = event not happening
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
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.
P(B)
P(A|B)
P(B')
P(A|B')
Conditional Formula
Where P(B) > 0
Multiplication Rule
= 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
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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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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