Real-Life Hook: Creating a Secure Password
You’re setting up a new email account. The password requirements are:
- First character: Must be a letter (26 options)
- Next 3 characters: Can be any digit (10 options each)
- Last character: Must be a special symbol (5 options)
How many different passwords can you create? This is where the Fundamental Principle of Counting becomes your superpower!
Answer: $26 \times 10 \times 10 \times 10 \times 5 = 130,000$ possible passwords.
Why This Topic Matters
The Fundamental Counting Principle is the foundation of all counting problems. Before you can master permutations and combinations, you need to understand when to multiply and when to add. This concept appears in:
- JEE Main: Direct questions (2-3 marks), foundational for P&C problems
- JEE Advanced: Building block for complex multi-step counting scenarios
- Real Life: Probability calculations, cryptography, decision trees, optimization
Core Concepts
1. Multiplication Principle (AND Rule)
Principle: If an event can occur in $m$ ways AND another independent event can occur in $n$ ways, then both events occurring together can happen in $m \times n$ ways.
$$\boxed{\text{Total ways} = m \times n \times p \times \ldots}$$Interactive Demo: Visualize Counting Principles
Explore how multiplication and addition principles work with decision trees.
Think: “AND means MULTIPLY” - you’re doing one thing AND then another.
Example: To reach a destination, you can go by:
- Train: 3 routes available
- Then bus: 4 routes available
- Total ways = $3 \times 4 = 12$ ways
2. Addition Principle (OR Rule)
Principle: If an event can occur in $m$ ways OR another mutually exclusive event can occur in $n$ ways, then either event can occur in $m + n$ ways.
$$\boxed{\text{Total ways} = m + n + p + \ldots}$$Think: “OR means ADD” - you’re choosing one option OR another (not both).
Example: To reach a destination, you can go by:
- Train: 3 routes available, OR
- Flight: 2 routes available
- Total ways = $3 + 2 = 5$ ways
3. Combined Principles
Most problems require both principles working together.
Example: College Admission
- You can choose Science stream (3 colleges) OR Commerce stream (4 colleges)
- After choosing stream AND college, you must choose a hostel (2 options)
Total ways = $(3 + 4) \times 2 = 14$ ways
Memory Tricks
AND vs OR - The Key Distinction
| Aspect | Multiplication (AND) | Addition (OR) |
|---|---|---|
| Keyword | “and then”, “followed by” | “or”, “alternatively” |
| Events | Sequential/simultaneous | Mutually exclusive choices |
| Operation | MULTIPLY | ADD |
| Result | Usually larger | Usually smaller |
| Example | Shirt AND pants | Bus OR train |
Mnemonic:
- M-AND = Multiply AND
- O-R = OR = Add (O looks like 0, + operation)
Decision Tree Technique
Draw a tree diagram when confused:
Start
├── Option 1 (m ways)
│ ├── Sub-option A (n ways)
│ └── Sub-option B (n ways)
└── Option 2 (p ways)
├── Sub-option A (n ways)
└── Sub-option B (n ways)
Total = m×n + p×n = n(m+p)
Common Counting Mistakes
❌ Mistake 1: Confusing AND with OR
Wrong: “Choose a shirt (5 types) and pants (3 types)” = $5 + 3 = 8$ ❌
Right: You need BOTH shirt AND pants = $5 \times 3 = 15$ ✓
Tip: Ask yourself - “Am I doing both or choosing one?”
❌ Mistake 2: Forgetting Independence
Problem: Two dice are thrown. Find total outcomes.
Wrong: Die 1 shows 6 values, Die 2 shows 6 values = $6 + 6 = 12$ ❌
Right: Each die is independent = $6 \times 6 = 36$ ✓
Tip: Independent events → Multiply; Mutually exclusive choices → Add
❌ Mistake 3: Not Considering Restrictions
Problem: 4-digit numbers using digits 1,2,3,4,5 (repetition allowed)
Common mistake: $5^4 = 625$ (includes numbers starting with 0… wait, we don’t have 0!)
Actually correct: $5^4 = 625$ ✓ (In this case, it’s fine!)
Real mistake scenario: If we included 0 in digits {0,1,2,3,4,5}, answer would be $5 \times 6^3$ not $6^4$
❌ Mistake 4: Over-counting
Problem: License plates with format: 2 letters then 3 digits
Wrong: “2 positions for letters, 3 positions for digits” = $2 \times 3 = 6$ ❌
Right: First letter (26) × Second letter (26) × First digit (10) × Second digit (10) × Third digit (10) = $26^2 \times 10^3$ ✓
Problem-Solving Framework
Step-by-Step Approach
- Identify: What needs to be counted?
- Break down: What are the independent steps/choices?
- Classify: Is each step AND (multiply) or OR (add)?
- Check restrictions: Any conditions limiting choices?
- Calculate: Apply principles systematically
- Verify: Does the answer make logical sense?
Practice Problems
Level 1: Foundation (JEE Main)
Problem 1.1: A person wants to go from city A to city C via city B. There are 4 routes from A to B and 3 routes from B to C. In how many ways can the person travel from A to C?
Solution
Answer: $4 \times 3 = 12$ ways
Explanation:
- Choose route A to B: 4 ways (AND)
- Choose route B to C: 3 ways
- Total = $4 \times 3 = 12$ ways
Problem 1.2: A restaurant offers:
- 3 types of starters
- 5 types of main course
- 2 types of desserts
If you must choose exactly one item from each category, how many different meals are possible?
Solution
Answer: $3 \times 5 \times 2 = 30$ different meals
Explanation: You choose one starter AND one main course AND one dessert (all three choices are independent and mandatory).
Problem 1.3: In a class, there are 10 boys and 8 girls. In how many ways can a class monitor (either a boy or a girl) be selected?
Solution
Answer: $10 + 8 = 18$ ways
Explanation: You select a boy OR a girl (mutually exclusive - can’t be both), so we add.
Problem 1.4: How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5 if: (a) Repetition is allowed? (b) Repetition is not allowed?
Solution
(a) With repetition: $5 \times 5 \times 5 = 125$
- Hundreds place: 5 choices
- Tens place: 5 choices
- Units place: 5 choices
(b) Without repetition: $5 \times 4 \times 3 = 60$
- Hundreds place: 5 choices
- Tens place: 4 remaining choices
- Units place: 3 remaining choices
Level 2: Intermediate (JEE Main/Advanced)
Problem 2.1: How many 4-letter codes can be formed using the letters of the word “EQUATION” if repetition is not allowed?
Solution
Answer: $8 \times 7 \times 6 \times 5 = 1680$
Explanation:
- Word “EQUATION” has 8 distinct letters
- First position: 8 choices
- Second position: 7 remaining
- Third position: 6 remaining
- Fourth position: 5 remaining
- Total = $8 \times 7 \times 6 \times 5 = 1680$
Problem 2.2: Five students compete for 3 prizes (first, second, third). In how many ways can the prizes be awarded if each student can receive at most one prize?
Solution
Answer: $5 \times 4 \times 3 = 60$ ways
Explanation:
- First prize: 5 choices
- Second prize: 4 remaining students
- Third prize: 3 remaining students
- Note: Order matters (first prize ≠ second prize)
Problem 2.3: A team consists of 5 boys and 4 girls. In how many ways can a team of 3 be selected such that: (a) All are boys? (b) All are girls? (c) At least one boy is included?
Solution
Wait! This requires combinations (nCr), which we’ll cover in later topics. Using fundamental counting:
Alternative approach for (c): Total ways - All girls = Total - (ways to select 3 girls)
We’ll solve this properly in the combinations chapter. For now, recognize this needs combination formulas, not just fundamental counting.
Learning point: Fundamental counting works best when order matters or when we’re making sequential choices. Selection problems often need combinations.
Problem 2.4: How many even numbers between 300 and 700 can be formed using digits 1, 2, 3, 4, 5, 6 without repetition?
Solution
Answer: 60
Explanation: Even numbers end in 2, 4, or 6. Numbers between 300 and 700 start with 3, 4, 5, or 6.
Case 1: Starts with 3, ends with even (2, 4, or 6)
- First position: 1 choice (3)
- Last position: 3 choices (2, 4, 6)
- Middle position: 4 remaining choices
- Total: $1 \times 4 \times 3 = 12$
Case 2: Starts with 4, ends with even (must be 2 or 6, not 4)
- First position: 1 choice (4)
- Last position: 2 choices (2, 6)
- Middle position: 4 remaining choices
- Total: $1 \times 4 \times 2 = 8$
Case 3: Starts with 5, ends with even (2, 4, or 6)
- Total: $1 \times 4 \times 3 = 12$
Case 4: Starts with 6, ends with even (must be 2 or 4, not 6)
- Total: $1 \times 4 \times 2 = 8$
Grand Total: $12 + 8 + 12 + 8 = 40$
Wait, let me recalculate more carefully:
For 3-digit number ABC where A is hundreds, B is tens, C is units:
- 300 ≤ ABC < 700
- A ∈ {3, 4, 5, 6}
- C ∈ {2, 4, 6} (for even numbers)
- No repetition
Case 1: A = 3
- C can be 2, 4, or 6 (3 choices)
- B can be any of remaining 4 digits (4 choices)
- Total: $3 \times 4 = 12$
Case 2: A = 4
- C must be 2 or 6 (2 choices, since 4 is used)
- B can be any of remaining 4 digits (4 choices)
- Total: $2 \times 4 = 8$
Case 3: A = 5
- C can be 2, 4, or 6 (3 choices)
- B can be any of remaining 4 digits (4 choices)
- Total: $3 \times 4 = 12$
Case 4: A = 6
- C must be 2 or 4 (2 choices, since 6 is used)
- B can be any of remaining 4 digits (4 choices)
- Total: $2 \times 4 = 8$
Answer: $12 + 8 + 12 + 8 = 40$
Level 3: Advanced (JEE Advanced)
Problem 3.1: In how many ways can 5 people be seated in a row such that two particular people are not adjacent?
Solution
Answer: 72
Explanation: Method 1: Complementary Counting
- Total arrangements of 5 people = $5! = 120$
- Arrangements where 2 particular people ARE adjacent:
- Treat them as one unit: 4 units to arrange = $4! = 24$
- Within the unit, 2 people can arrange in $2! = 2$ ways
- Total with adjacency = $24 \times 2 = 48$
- Answer = $120 - 48 = 72$
Method 2: Direct Counting We’ll learn this method with permutations!
Problem 3.2: A 5-digit number is formed using digits 0, 1, 2, 3, 4, 5 without repetition. How many such numbers are divisible by 4?
Solution
Divisibility rule for 4: Last two digits must form a number divisible by 4.
From {0,1,2,3,4,5}, two-digit numbers divisible by 4:
- 04, 12, 20, 24, 32, 40, 52
But for 5-digit numbers:
- First digit ≠ 0
- All digits distinct
This requires careful case analysis:
Last two digits = 04:
- First digit: 4 choices {1,2,3,5}
- Second digit: 3 remaining
- Third digit: 2 remaining
- Count: $4 \times 3 \times 2 = 24$
Last two digits = 12:
- Remaining digits: {0,3,4,5}
- First digit: 3 choices {3,4,5} (not 0)
- Second digit: 3 remaining (including 0 now)
- Third digit: 2 remaining
- Count: $3 \times 3 \times 2 = 18$
Last two digits = 20:
- Remaining: {1,3,4,5}
- First: 4 choices, Second: 3, Third: 2
- Count: $4 \times 3 \times 2 = 24$
Last two digits = 24:
- Remaining: {0,1,3,5}
- First: 3 choices {1,3,5}, Second: 3, Third: 2
- Count: $3 \times 3 \times 2 = 18$
Last two digits = 32:
- Remaining: {0,1,4,5}
- First: 3 choices {1,4,5}, Second: 3, Third: 2
- Count: $3 \times 3 \times 2 = 18$
Last two digits = 40:
- Remaining: {1,2,3,5}
- First: 4, Second: 3, Third: 2
- Count: $4 \times 3 \times 2 = 24$
Last two digits = 52:
- Remaining: {0,1,3,4}
- First: 3 choices {1,3,4}, Second: 3, Third: 2
- Count: $3 \times 3 \times 2 = 18$
Total: $24 + 18 + 24 + 18 + 18 + 24 + 18 = 144$
Problem 3.3: Find the number of ways to arrange the letters of the word “MATHEMATICS” (we’ll revisit this with permutations, but try using fundamental counting).
Solution
Letters in MATHEMATICS: M(2), A(2), T(2), H(1), E(1), I(1), C(1), S(1)
- Total 11 letters
- Repeated: M, A, T (each appears twice)
Using fundamental counting is complex here. We need:
$$\frac{11!}{2! \times 2! \times 2!}$$Why? This requires understanding of permutations with repetition (next topic!).
Learning point: Fundamental counting has limits. When dealing with repetitions and arrangements, permutation formulas are more efficient.
Problem 3.4: A password consists of 8 characters. The first 2 must be uppercase letters, next 4 must be digits, and last 2 must be lowercase letters. Characters can repeat. However, the 4 digits must include at least one even digit. How many such passwords are possible?
Solution
Answer: $26^2 \times (10^4 - 5^4) \times 26^2$
Explanation:
- First 2 positions (uppercase): $26 \times 26 = 26^2$
- Last 2 positions (lowercase): $26 \times 26 = 26^2$
- Middle 4 positions (digits with at least one even):
- Total 4-digit combinations: $10^4$
- All odd digits (no even): $5^4$ (digits 1,3,5,7,9)
- At least one even: $10^4 - 5^4$
Total: $26^2 \times (10000 - 625) \times 26^2 = 676 \times 9375 \times 676$
$= 4,289,250,000$ passwords
Technique: Complementary counting (Total - Unwanted)
Cross-Topic Connections
Link to Permutations
The fundamental counting principle is the foundation of permutations. When you calculate $nPr = \frac{n!}{(n-r)!}$, you’re actually using multiplication principle:
- Choose 1st object: $n$ ways
- Choose 2nd object: $(n-1)$ ways
- …
- Choose $r$-th object: $(n-r+1)$ ways
- Total: $n \times (n-1) \times \ldots \times (n-r+1) = nPr$
→ See Permutations Basics
Link to Combinations
Addition principle helps when combining different cases in selection problems.
→ See Combinations Basics
Link to Probability
$$P(\text{Event}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}}$$Both numerator and denominator are often calculated using counting principles!
→ See Probability
Link to Binomial Theorem
The number of terms in binomial expansion $(x+y)^n$ uses counting principles.
→ See Binomial Theorem
JEE Tips & Tricks
Time-Saving Strategies
- Draw diagrams for complex problems (decision trees, slot method)
- Use complementary counting when “at least” or “at most” appears
- Check answer reasonableness: Does $5 + 3 = 15$ make sense? No! Should be $5 \times 3$.
- Slot method: Draw blanks
_ _ _and fill each position
Common JEE Patterns
- License plates: Letters and digits with restrictions
- Passwords: Character types with conditions
- Teams/Committees: Selection with constraints (→ combinations)
- Arrangement problems: Seating, standing (→ permutations)
- Divisibility: Numbers with specific properties
Quick Checklist Before Solving
- What exactly am I counting?
- Is it sequential (AND) or alternatives (OR)?
- Any restrictions on choices?
- Can choices repeat?
- Does order matter? (If yes → likely permutation later)
Summary
| Principle | Operation | Keyword | Example |
|---|---|---|---|
| Multiplication | × | “AND”, “then” | Shirt AND pants: $5 \times 3$ |
| Addition | + | “OR”, “alternatively” | Bus OR train: $3 + 2$ |
Key Takeaways:
- Multiplication for sequential/simultaneous independent events
- Addition for mutually exclusive choices
- Most problems combine both principles
- Always check for restrictions and conditions
- Draw diagrams when confused!
Next Steps:
- Master Permutations Basics to handle ordered arrangements
- Learn Combinations for unordered selections
Last updated: September 5, 2025