Permutation & Combination - Theory & Concepts

🔀 Permutation & Combination - Complete Theory

Master counting, arrangement, and selection problems!


🎯 Key Difference

Permutation (Order Matters!)

Arrangement of objects where ORDER is important.

Example: ABC, ACB, BAC, BCA, CAB, CBA
All are DIFFERENT permutations!

Symbol: ⁿPᵣ or P(n,r)

Combination (Order Doesn’t Matter!)

Selection of objects where ORDER is NOT important.

Example: ABC, ACB, BAC, BCA, CAB, CBA
All are the SAME combination!

Symbol: ⁿCᵣ or C(n,r) or (n choose r)

Memory Trick:

  • Permutation → Position matters
  • Combination → Choice/Collection (order irrelevant)

📐 Basic Formulas

Factorial

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

0! = 1 (by definition)
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = 5,040
8! = 40,320
9! = 362,880
10! = 3,628,800

Permutation Formula

ⁿPᵣ = n! / (n-r)!

Where:
n = total number of objects
r = objects to arrange

Example: ⁵P₃ = 5! / (5-3)!
              = 5! / 2!
              = 120 / 2
              = 60

Special Case:

ⁿPₙ = n! (arranging all n objects)
⁵P₅ = 5! = 120

Combination Formula

ⁿCᵣ = n! / [r! × (n-r)!]

Where:
n = total number of objects
r = objects to select

Example: ⁵C₃ = 5! / [3! × 2!]
              = 120 / (6 × 2)
              = 120 / 12
              = 10

Important Properties:

ⁿC₀ = 1
ⁿCₙ = 1
ⁿCᵣ = ⁿCₙ₋ᵣ
ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ

🔄 Relationship Between P & C

ⁿPᵣ = ⁿCᵣ × r!

Permutation = Combination × Arrangement of selected items

Example: ⁵P₃ = ⁵C₃ × 3!
              = 10 × 6
              = 60 ✓

💡 Solved Examples

Example 1: Basic Permutation

Q: How many 3-digit numbers can be formed using digits 1, 2, 3, 4, 5 without repetition?

Solution:

We need to arrange 3 digits from 5
Order matters (123 ≠ 321)

⁵P₃ = 5! / 2! = 120 / 2 = 60

Answer: 60


Example 2: Basic Combination

Q: In how many ways can 3 students be selected from 5 students?

Solution:

Selection, order doesn't matter

⁵C₃ = 5! / (3! × 2!)
    = 120 / (6 × 2)
    = 10

Answer: 10


Example 3: Arranging Letters

Q: How many ways can letters of “BOOK” be arranged?

Solution:

Total letters = 4
But O appears 2 times (identical)

Arrangements = 4! / 2!
             = 24 / 2
             = 12

Answer: 12


Example 4: Circular Permutation

Q: 5 people sitting around circular table. Find number of arrangements.

Solution:

Circular permutation = (n-1)!

= (5-1)!
= 4!
= 24

Answer: 24


Example 5: Selecting with Restriction

Q: From 5 boys and 4 girls, select committee of 3 boys and 2 girls. How many ways?

Solution:

Boys: ⁵C₃ = 5!/(3!×2!) = 10
Girls: ⁴C₂ = 4!/(2!×2!) = 6

Total = 10 × 6 = 60

Answer: 60


Example 6: At Least/At Most

Q: From 6 people, committees of at least 2 can be formed in how many ways?

Solution:

At least 2 = 2 or 3 or 4 or 5 or 6

= ⁶C₂ + ⁶C₃ + ⁶C₄ + ⁶C₅ + ⁶C₆
= 15 + 20 + 15 + 6 + 1
= 57

Shortcut: Total - unwanted
= 2⁶ - ⁶C₀ - ⁶C₁
= 64 - 1 - 6
= 57 ✓

Answer: 57


Example 7: Identical Objects

Q: How many ways to arrange letters of “MISSISSIPPI”?

Solution:

Total = 11 letters
M = 1, I = 4, S = 4, P = 2

Arrangements = 11! / (1! × 4! × 4! × 2!)
             = 39,916,800 / (1 × 24 × 24 × 2)
             = 34,650

Answer: 34,650


Example 8: Specific Positions

Q: Word “COMPUTER”. How many arrangements start with C and end with R?

Solution:

C _ _ _ _ _ _ R

C and R are fixed
Remaining 6 letters to arrange in middle

= 6!
= 720

Answer: 720


🎯 Important Patterns

Pattern 1: Arrangements of n Items

Total arrangements = n!

But if some are identical:
= n! / (p! × q! × r! × ...)

where p, q, r are frequencies of identical items

Pattern 2: Circular Arrangements

Linear: n!
Circular: (n-1)!

If clockwise = anticlockwise (like beads):
= (n-1)! / 2

Pattern 3: Selecting All vs Some

From n items:
Select all: Only 1 way (ⁿCₙ = 1)
Arrange all: n! ways

Select r: ⁿCᵣ ways
Arrange r: ⁿPᵣ ways

Pattern 4: Vowels/Consonants Together

Treat grouped items as single unit

Example: ORANGE, keep vowels together
Vowels: OAE (treat as 1 unit)
Units: (OAE), R, N, G = 4 units

Arrangements = 4! × 3!
(4! for units, 3! for vowels within)

⚡ Quick Shortcuts

Shortcut 1: ⁿCᵣ Calculation

⁷C₃ = (7 × 6 × 5) / (3 × 2 × 1)
    = 210 / 6
    = 35

Easier than calculating full factorials!

Shortcut 2: Symmetry Property

ⁿCᵣ = ⁿCₙ₋ᵣ

If calculating ²⁰C₁₈ is hard:
²⁰C₁₈ = ²⁰C₂ = 190 (much easier!)

Shortcut 3: Consecutive Selection

Selecting r consecutive items from n items in circle:
= n ways

Example: 3 consecutive from 10 in circle = 10

Shortcut 4: Divisibility by Specific Digit

To count numbers divisible by 5:
Last digit must be 0 or 5
Fix last digit, arrange remaining

📊 Special Cases

Selecting from Groups

From m men and n women, select:
- r men and s women: ᵐCᵣ × ⁿCₛ
- at least 1 man: Total - (all women)
- exactly 1 woman: ⁿC₁ × ᵐCᵣ₋₁

Distributing Items

Distributing n identical items to r people:
Each gets at least 1: ⁿ⁻¹Cᵣ₋₁
No restriction: ⁿ⁺ʳ⁻¹Cᵣ₋₁

Derangements

Number of ways to arrange n items where no item is in its original position:
≈ n! / e

⚠️ Common Mistakes

❌ Mistake 1: Confusing P & C

Wrong: Using combination when order matters ✗
Right:
  Arrangement/Ranking/Code → Permutation ✓
  Selection/Committee/Team → Combination ✓

❌ Mistake 2: Identical Items

Wrong: Counting "BOOK" as 4! = 24 ✗
Right: 4!/2! = 12 (two O's are identical) ✓

❌ Mistake 3: Circular ≠ Linear

Wrong: 5 people in circle = 5! ✗
Right: = (5-1)! = 4! ✓

❌ Mistake 4: At Least Calculation

Wrong: Direct addition (tedious, error-prone) ✗
Right: Total - Unwanted (complement!) ✓

❌ Mistake 5: Selection Formula

Wrong: ⁿCᵣ = n! / r! ✗
Right: ⁿCᵣ = n! / (r! × (n-r)!) ✓

📝 Practice Problems

Level 1:

  1. How many 4-digit numbers from digits 1-9 without repetition?
  2. In how many ways can 5 books be arranged on shelf?
  3. From 10 people, select committee of 4. How many ways?

Level 2:

  1. How many ways to arrange letters of “SUCCESS”?
  2. 6 people around circular table. Find arrangements.
  3. From 8 boys and 6 girls, select 3 boys and 2 girls. How many ways?

Level 3:

  1. Word “EQUATION”. Arrangements starting with vowel?
  2. 10 points, no 3 collinear. How many triangles?
  3. Distribute 12 identical chocolates to 5 children, each gets at least 1. How many ways?

Uses P&C:

Related:

Practice:


Master P&C - Ask: Does order matter? 🔀