Real-Life Hook: The Podium Problem
Imagine an Olympic race with 8 athletes. In how many different ways can the Gold, Silver, and Bronze medals be awarded?
Think about it:
- If athlete A wins gold, there are 7 choices for silver, then 6 for bronze
- If athlete B wins gold, there are 7 choices for silver, then 6 for bronze
- …and so on for all 8 athletes
Answer: $8 \times 7 \times 6 = 336$ ways
This is a permutation problem: $^8P_3 = 336$
Interactive Demo: Visualize Permutations
See how order matters in arrangements with probability trees.
Key insight: Order matters! Gold-Silver-Bronze is different from Bronze-Silver-Gold.
Why This Topic Matters
Permutations are fundamental to counting ordered arrangements. This is one of the most frequently tested topics in JEE:
- JEE Main: 3-5 questions (direct and indirect), 12-15 marks
- JEE Advanced: Complex multi-step problems, often combined with restrictions
- Real Life: Scheduling, ranking systems, playlist order, password sequences
Critical distinction: Permutation (order matters) vs Combination (order doesn’t matter)
- Arranging 3 books: Permutation (ABC ≠ BAC)
- Selecting 3 books: Combination (choosing {A,B,C} = choosing {C,B,A})
Interactive Demo: Permutation Calculator
Permutation (nPr) Calculator
Calculate nPr = n!/(n-r)! - the number of ways to arrange r objects from n distinct objects
Core Concepts
1. Factorial Notation
Definition: For a positive integer $n$:
$$\boxed{n! = n \times (n-1) \times (n-2) \times \ldots \times 3 \times 2 \times 1}$$Special cases:
- $\boxed{0! = 1}$ (by definition, crucial for formulas)
- $\boxed{1! = 1}$
Examples:
- $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
- $3! = 6$
- $10! = 3,628,800$
Recursive property: $n! = n \times (n-1)!$
Memory trick: Factorial grows EXTREMELY fast!
- $10! = 3.6$ million
- $20! = 2.4$ quintillion (19 digits!)
- Never calculate large factorials directly; simplify first!
2. Permutation of n Distinct Objects
Question: In how many ways can $n$ distinct objects be arranged in a row?
Answer:
$$\boxed{n!}$$Reasoning:
- First position: $n$ choices
- Second position: $(n-1)$ remaining choices
- Third position: $(n-2)$ remaining choices
- …
- Last position: $1$ choice
- Total: $n \times (n-1) \times (n-2) \times \ldots \times 1 = n!$
Example: Arranging 4 books A, B, C, D
- Total arrangements = $4! = 24$
3. Permutation Formula: nPr
Question: In how many ways can we select and arrange $r$ objects from $n$ distinct objects?
Formula:
$$\boxed{^nP_r = \frac{n!}{(n-r)!} = n \times (n-1) \times (n-2) \times \ldots \times (n-r+1)}$$Alternative notation: $P(n,r)$ or $P_r^n$ or $_nP_r$
Reasoning:
- Choose 1st object: $n$ ways
- Choose 2nd object: $(n-1)$ ways
- …
- Choose $r$-th object: $(n-r+1)$ ways
- Total: $n \times (n-1) \times \ldots \times (n-r+1)$
Why the formula works:
$$^nP_r = n \times (n-1) \times \ldots \times (n-r+1) = \frac{n!}{(n-r)!}$$Example: From 8 athletes, ways to award 3 medals:
$$^8P_3 = \frac{8!}{5!} = \frac{8!}{5!} = 8 \times 7 \times 6 = 336$$Special cases:
- $^nP_0 = 1$ (selecting and arranging 0 objects = 1 way: do nothing)
- $^nP_1 = n$ (selecting and arranging 1 object = $n$ ways)
- $^nP_n = n!$ (selecting and arranging all $n$ objects)
- $^nP_r = 0$ if $r > n$ (can’t select more than available)
4. Important Properties
$$\boxed{^nP_r = n \times ^{n-1}P_{r-1}}$$Proof:
$$^nP_r = \frac{n!}{(n-r)!} = n \times \frac{(n-1)!}{(n-r)!} = n \times \frac{(n-1)!}{((n-1)-(r-1))!} = n \times ^{n-1}P_{r-1}$$ $$\boxed{\frac{^nP_r}{^nP_{r-1}} = n - r + 1}$$Proof:
$$\frac{^nP_r}{^nP_{r-1}} = \frac{n!/(n-r)!}{n!/(n-r+1)!} = \frac{(n-r+1)!}{(n-r)!} = n-r+1$$Memory Tricks
nPr vs nCr - The Ultimate Confusion Buster
| Aspect | Permutation (nPr) | Combination (nCr) |
|---|---|---|
| Order | Matters | Doesn’t matter |
| Formula | $\frac{n!}{(n-r)!}$ | $\frac{n!}{r!(n-r)!}$ |
| Relation | $nPr = nCr \times r!$ | $nCr = \frac{nPr}{r!}$ |
| Example | ABC ≠ BAC | {A,B,C} = {B,A,C} |
| Keywords | “Arrange”, “order”, “sequence” | “Select”, “choose”, “group” |
| Real-life | Password, race ranking | Team selection, lottery |
Mnemonic for nPr:
- P = Position matters
- P = Podium (ranks: 1st, 2nd, 3rd)
Mnemonic for nCr:
- C = Choose (selection only)
- C = Committee (no rank within group)
Factorial Simplification Trick
Never expand both numerator and denominator!
$$\frac{10!}{7!} = \frac{10 \times 9 \times 8 \times \cancel{7!}}{\cancel{7!}} = 10 \times 9 \times 8 = 720$$NOT: $\frac{3,628,800}{5,040} = 720$ (painful calculation!)
Quick Factorial Values
Memorize these:
- $0! = 1$
- $1! = 1$
- $2! = 2$
- $3! = 6$
- $4! = 24$
- $5! = 120$
- $6! = 720$
- $7! = 5,040$
- $10! = 3,628,800$
Common Counting Mistakes
❌ Mistake 1: Confusing Permutation with Combination
Problem: Choose 3 students from 10 to form a team.
Wrong: $^{10}P_3 = 720$ ❌ (This counts ordered selections!)
Right: $^{10}C_3 = 120$ ✓ (Team has no order/rank)
How to identify: Does ABC team = BAC team? YES → Combination. NO → Permutation.
❌ Mistake 2: Forgetting 0! = 1
Problem: Find $^5P_5$
Wrong: $\frac{5!}{(5-5)!} = \frac{5!}{0}$ = undefined ❌
Right: $\frac{5!}{0!} = \frac{120}{1} = 120 = 5!$ ✓
Why 0! = 1? By definition, and it makes formulas consistent. There’s exactly 1 way to arrange 0 objects (do nothing).
❌ Mistake 3: Not Simplifying Before Calculating
Problem: Calculate $\frac{20!}{17!}$
Wrong: Calculate $20! = 2.43 \times 10^{18}$, then calculate $17! = 3.56 \times 10^{14}$, then divide ❌
Right: $\frac{20!}{17!} = 20 \times 19 \times 18 = 6,840$ ✓
❌ Mistake 4: Ignoring Restrictions
Problem: Arrange 5 people in a row such that A and B must be together.
Wrong: $5! = 120$ ❌ (Counts all arrangements, including A and B separated)
Right: Treat AB as one unit → $4! \times 2! = 48$ ✓
Problem-Solving Strategies
Strategy 1: Slot Method
Draw blanks for positions, fill each position considering restrictions.
Example: 4-digit numbers from {1,2,3,4,5,6} without repetition
_ _ _ _
↓ ↓ ↓ ↓
6 5 4 3 = 6P4 = 360
Strategy 2: Treat Groups as Single Units
When objects must be together, bundle them.
Example: ABCDE where AB must be together
- Units: (AB), C, D, E → 4 units
- Arrangements: $4!$
- Within (AB): $2!$ ways
- Total: $4! \times 2! = 48$
Strategy 3: Complementary Counting
Total arrangements - Restricted arrangements
Example: 5 people where A and B are NOT together
- Total: $5! = 120$
- A and B together: $4! \times 2! = 48$
- A and B NOT together: $120 - 48 = 72$
Strategy 4: Fix One Object
When dealing with circular or relative arrangements, fix one object to remove rotational duplicates.
Practice Problems
Level 1: Foundation (JEE Main)
Problem 1.1: Find the value of $^7P_3$.
Solution
Answer: 210
Method 1 (Direct formula):
$$^7P_3 = \frac{7!}{(7-3)!} = \frac{7!}{4!} = 7 \times 6 \times 5 = 210$$Method 2 (Multiplication):
$$^7P_3 = 7 \times 6 \times 5 = 210$$Problem 1.2: In how many ways can 6 different books be arranged on a shelf?
Solution
Answer: $6! = 720$
Explanation: Arranging $n$ distinct objects = $n!$
Problem 1.3: How many 3-letter words (with or without meaning) can be formed using the letters of the word “NUMERIC” without repetition?
Solution
Answer: 210
Explanation:
- Word “NUMERIC” has 7 distinct letters
- We need to select and arrange 3 letters
- Total ways = $^7P_3 = 7 \times 6 \times 5 = 210$
Problem 1.4: Find $n$ if $^nP_2 = 20$.
Solution
Answer: $n = 5$
Solution:
$$^nP_2 = n(n-1) = 20$$ $$n^2 - n - 20 = 0$$ $$(n-5)(n+4) = 0$$ $$n = 5 \text{ or } n = -4$$Since $n$ must be positive, $n = 5$.
Verification: $^5P_2 = 5 \times 4 = 20$ ✓
Level 2: Intermediate (JEE Main/Advanced)
Problem 2.1: In how many ways can 5 boys and 3 girls be arranged in a row such that no two girls are together?
Solution
Answer: 14,400
Strategy: First arrange boys, then place girls in gaps.
Step 1: Arrange 5 boys in a row
- Ways = $5! = 120$
Step 2: Create positions for girls
_B_B_B_B_B_
6 positions (gaps) created for 3 girls
Step 3: Choose 3 positions from 6 gaps and arrange girls
- Ways = $^6P_3 = 6 \times 5 \times 4 = 120$
Total: $120 \times 120 = 14,400$
Alternative thinking:
- Choose 3 gaps from 6: $^6P_3$ (order matters because girls are distinct)
- Arrange boys: $5!$
- Total: $5! \times ^6P_3$
Problem 2.2: Find the number of ways in which the letters of the word “MONDAY” can be arranged so that the words thus formed begin with M and end with Y.
Solution
Answer: 24
Explanation:
- First position: M (fixed)
- Last position: Y (fixed)
- Middle 4 positions: O, N, D, A can be arranged in $4!$ ways
Total: $4! = 24$
Problem 2.3: If $^{10}P_r = 5040$, find $r$.
Solution
Answer: $r = 4$
Solution:
$$^{10}P_r = \frac{10!}{(10-r)!} = 5040$$Try values:
- $^{10}P_1 = 10$
- $^{10}P_2 = 90$
- $^{10}P_3 = 720$
- $^{10}P_4 = 5040$ ✓
Alternatively, recognize that $7! = 5040$:
$$\frac{10!}{(10-r)!} = 7!$$ $$\frac{10!}{(10-r)!} = 7!$$ $$10 \times 9 \times 8 \times 7! = 7! \times (10-r)!$$This gives us:
$$10 \times 9 \times 8 \times 7 = \frac{10!}{6!}$$ $$^{10}P_4 = 5040$$So $r = 4$.
Problem 2.4: In how many ways can 8 people be seated in a row if: (a) There are no restrictions? (b) Two specific people must sit together? (c) Two specific people must NOT sit together?
Solution
(a) No restrictions: $8! = 40,320$
(b) Two people must sit together:
- Treat 2 people as 1 unit: 7 units total
- Arrange 7 units: $7! = 5,040$
- Within the unit, 2 people can arrange: $2! = 2$
- Total: $7! \times 2! = 10,080$
(c) Two people must NOT sit together:
- Total - (Two together)
- $8! - 7! \times 2! = 40,320 - 10,080 = 30,240$
Level 3: Advanced (JEE Advanced)
Problem 3.1: Find the number of ways to arrange the letters of “GARDEN” such that vowels occupy even positions.
Solution
Answer: 144
Analysis:
- Word: G, A, R, D, E, N (6 letters)
- Vowels: A, E (2 vowels)
- Consonants: G, R, D, N (4 consonants)
- Positions: 1, 2, 3, 4, 5, 6
- Even positions: 2, 4, 6 (3 positions)
- Odd positions: 1, 3, 5 (3 positions)
Step 1: Arrange 2 vowels in 3 even positions
- Ways = $^3P_2 = 3 \times 2 = 6$
Step 2: Arrange 4 consonants in remaining 4 positions (3 odd + 1 even)
- Ways = $4! = 24$
Total: $6 \times 24 = 144$
Problem 3.2: How many numbers between 3000 and 4000 can be formed using digits 0, 1, 2, 3, 4, 5, 6 without repetition?
Solution
Answer: 120
Analysis:
- Numbers between 3000 and 4000: Form is 3ABC
- First digit: Must be 3 (1 choice)
- Remaining digits: Choose from {0,1,2,4,5,6} (6 digits)
- Second, third, fourth positions: Select and arrange 3 from 6
Calculation:
- First position: 1 way (digit 3)
- Remaining 3 positions: $^6P_3 = 6 \times 5 \times 4 = 120$
Total: $1 \times 120 = 120$
Problem 3.3: In how many ways can 5 men and 5 women be seated in a row so that men and women alternate?
Solution
Answer: 28,800
Analysis: Two patterns possible:
- M W M W M W M W M W
- W M W M W M W M W M
Pattern 1 (starts with man):
- Men in positions 1,3,5,7,9: $5!$ ways
- Women in positions 2,4,6,8,10: $5!$ ways
- Total: $5! \times 5! = 120 \times 120 = 14,400$
Pattern 2 (starts with woman):
- Women in odd positions: $5!$ ways
- Men in even positions: $5!$ ways
- Total: $5! \times 5! = 14,400$
Grand Total: $14,400 + 14,400 = 28,800$
Problem 3.4: Find the rank of the word “SNAKE” when all letters are arranged in alphabetical order.
Solution
Answer: 106
Solution:
First, alphabetically sort letters: A, E, K, N, S
Words before SNAKE:
Starting with A: AEKNS, AEKSN, AENKS, AENSK, AESNK, … (all arrangements of EKNS)
- Count: $4! = 24$
Starting with E: EAKNS, EAKSN, … (all arrangements of AKNS)
- Count: $4! = 24$
Starting with K: KAENS, KAESN, … (all arrangements of AENS)
- Count: $4! = 24$
Starting with N: NAEKS, NAESK, … (all arrangements of AEKS)
- Count: $4! = 24$
Starting with SA: SAEKN, SAENK, SAKNE, … (all arrangements starting with SA)
- Count: $3! = 6$
Starting with SE: SEAKN, SEANK, SEKNA, … (all arrangements starting with SE)
- Count: $3! = 6$
Starting with SK: SKANE, SKAEN, … (all arrangements starting with SK)
- Count: $3! = 6$
Starting with SNA: SNAEK, SNAKE, SNEAK, …
Before SNAKE starting with SNA:
- SNAEK (comes before SNAKE)
- Count: 1
Total words before SNAKE: $24 + 24 + 24 + 24 + 6 + 6 + 6 + 1 = 115$
Rank of SNAKE: $115 + 1 = 116$
Wait, let me recalculate more carefully.
Alphabetical order: A, E, K, N, S
Starting with A: $4! = 24$ words Starting with E: $4! = 24$ words Starting with K: $4! = 24$ words Starting with N: $4! = 24$ words
So far: 96 words
Starting with S:
- Starting with SA: $3! = 6$ words
- Starting with SE: $3! = 6$ words
- Starting with SK: $3! = 6$ words
- Starting with SN:
- Starting with SNA:
- SNAEK (E before K)
- SNAKE ← This is our word!
- Starting with SNA:
Total before SNAKE: $96 + 6 + 6 + 6 + 1 = 115$
Rank of SNAKE = $115 + 1 = 116$
Actually, I made an error. Let me recalculate:
Starting with SN:
- SNA:
- SNAEK
- SNAKE
- SNE:
- SNEAK
- SNAEKA (not valid)
Total before SNAKE in SNA_: just SNAEK, so 1 word.
Total: $96 + 18 + 1 = 115$
Rank = 116
Let me verify once more systematically:
Words starting with:
- A____: $4! = 24$
- E____: $4! = 24$
- K____: $4! = 24$
- N____: $4! = 24$
Total: 96
Words starting with S:
- SA___: $3! = 6$
- SE___: $3! = 6$
- SK___: $3! = 6$
Total so far: $96 + 18 = 114$
Words starting with SN:
- SNAEK: 1
- SNAKE: This is our word!
Total before SNAKE: $114 + 1 = 115$
Rank of SNAKE = 116
Hmm, I’m getting 116. Let me reconsider the problem once more to be sure.
Actually, I realize I need to be more careful. Let me redo this:
Letters in alphabetical order: A, E, K, N, S
Count words before “SNAKE”:
- Words starting with A: $4! = 24$
- Words starting with E: $4! = 24$
- Words starting with K: $4! = 24$
- Words starting with N: $4! = 24$
Running total: 96 words
- Words starting with SA: $3! = 6$
- Words starting with SE: $3! = 6$
- Words starting with SK: $3! = 6$
Running total: $96 + 18 = 114$ words
- Words starting with SN and third letter A:
- SNAEK: 1 word
- SNAKE: Our target word!
Total before SNAKE: $114 + 1 = 115$ words
Rank of SNAKE: $115 + 1 = 116$
Wait, the answer I stated initially was 106. Let me reconsider…
Actually, I think I need to recalculate. The error might be in my counting. Let me be very systematic:
Alphabetical: A < E < K < N < S
Words starting with:
- First letter A: Remaining EKNS, arrangements = $4! = 24$
- First letter E: Remaining AKNS, arrangements = $4! = 24$
- First letter K: Remaining AENS, arrangements = $4! = 24$
- First letter N: Remaining AEKS, arrangements = $4! = 24$
Subtotal: 96 words
Words starting with S (target starts with S):
- Second letter A (SA___): Remaining EKN, arrangements = $3! = 6$
- Second letter E (SE___): Remaining AKN, arrangements = $3! = 6$
- Second letter K (SK___): Remaining AEN, arrangements = $3! = 6$
Subtotal: $96 + 18 = 114$ words
Words starting with SN (target is SN___):
- Third letter A (SNA__): Remaining EK
- SNAEK (E before K)
- SNAKE (K before E) ← TARGET
Before SNAKE starting with SNA: 1 word (SNAEK)
Total before SNAKE: $114 + 1 = 115$
Rank of SNAKE: 116
So the correct answer should be 116, not 106. I’ll correct this.
Answer: 116
Cross-Topic Connections
Link to Combinations
Permutation and combination are related:
$$\boxed{^nP_r = ^nC_r \times r!}$$Why?
- $^nC_r$ = Ways to select $r$ objects (no order)
- $r!$ = Ways to arrange those $r$ objects
- $^nP_r$ = Select AND arrange = $^nC_r \times r!$
→ See Combinations Basics
Link to Circular Permutations
Arranging objects in a circle is different from a line due to rotational symmetry.
→ See Circular Permutations
Link to Probability
$$P = \frac{\text{Favorable permutations}}{\text{Total permutations}}$$Many probability problems involve counting permutations.
→ See Probability
Link to Binomial Theorem
Expansion of $(x+y)^n$ has $^nC_r$ coefficients, which relates to permutations:
$$^nC_r = \frac{^nP_r}{r!}$$→ See Binomial Theorem
JEE Tips & Tricks
Time-Saving Strategies
- Never calculate large factorials: Always simplify $\frac{n!}{m!}$ by canceling
- Use $^nP_r = n \times (n-1) \times \ldots \times (n-r+1)$: Faster than formula with factorials
- Complementary counting: When restrictions are complex, use Total - Unwanted
- Slot method: Draw blanks, fill systematically
Common JEE Patterns
- Rank of a word: Systematic counting in alphabetical order
- Restricted arrangements: Adjacent/non-adjacent, specific positions
- Equation solving: Given $^nP_r$, find $n$ or $r$
- Mixed problems: Combine permutations with other topics
Quick Checklist
- Does order matter? (YES → Permutation, NO → Combination)
- Are all objects distinct? (If not → Permutations with repetition)
- Any restrictions? (Together, apart, specific positions?)
- Simplify before calculating!
Summary Table
| Concept | Formula | Example |
|---|---|---|
| Factorial | $n! = n \times (n-1) \times \ldots \times 1$ | $5! = 120$ |
| Arrange n objects | $n!$ | Arrange 6 books = $6!$ |
| Select & arrange r from n | $^nP_r = \frac{n!}{(n-r)!}$ | 3 medals from 8 = $^8P_3$ |
| Relation with nCr | $^nP_r = ^nC_r \times r!$ | $^5P_2 = ^5C_2 \times 2!$ |
Key Takeaway: Permutations count ordered arrangements. When order matters, use nPr!
Next Steps:
- Learn Permutations with Repetition for repeated objects
- Master Circular Permutations for circular arrangements
- Understand Combinations for unordered selections
Last updated: September 8, 2025