Permutation and Combination Explained with Formulas and Examples - Aptitude Questions & Answers
Permutation and Combination Explained with Formulas and Examples is one of the most important topics in Quantitative Aptitude. In this lesson, you will learn concepts, formulas, shortcuts, solved examples, and aptitude questions with answers. This topic is useful for exams like SSC, Bank, CAT, TCS, and other competitive exams.
Permutations and Combinations: Complete Guide with Formulas & Examples
What are Permutations and Combinations?
Permutations and Combinations are fundamental counting principles in mathematics used to calculate the number of possible arrangements and selections of objects. These concepts are crucial for probability, statistics, and various competitive exams.
Key Difference: Permutation considers arrangement order, while Combination does not.
Fundamental Counting Principle
If one event can occur in \'m\' ways and another independent event can occur in \'n\' ways, then the two events together can occur in m × n ways.
Example: If you have 3 shirts and 4 pants, you can create 3 × 4 = 12 different outfits.
Permutation - Arrangement with Order
Definition and Formula
Permutation refers to the arrangement of objects in a specific order. The order of arrangement matters in permutation.
Where:
n = Total number of objects
r = Number of objects to arrange
! = Factorial (e.g., 5! = 5×4×3×2×1 = 120)
Example: In how many ways can 8 people sit in 8 vacant chairs?
Number of ways = 8P8 = 8! = 8×7×6×5×4×3×2×1 = 40,320 ways
Explanation: First chair: 8 choices, Second: 7 choices, Third: 6 choices, and so on.
Types of Permutation Problems
1. Number-Based Permutation
With Repetition Allowed
Each digit/number can be used multiple times in different positions.
Total permutations = m × m × m ... (n times) = mⁿ
Example Problem: How many 4-digit numbers can be formed using digits 1,5,3,4,0,6 if repetition is allowed?
- Available digits: {1,5,3,4,0,6} → 6 digits total
- First digit cannot be 0 → 5 choices (1,5,3,4,6)
- Other positions can be any digit → 6 choices each
- Total = 5 × 6 × 6 × 6 = 1,080 numbers
Without Repetition
Each digit/number can be used only once in the arrangement.
Total permutations = mPₙ = m!/(m-n)!
Example Problem: How many 5-digit numbers can be formed using digits 1,4,7,8,3,6 without repetition?
- Available digits: {1,4,7,8,3,6} → 6 digits total
- First position: 6 choices
- Second position: 5 choices (one digit used)
- Third position: 4 choices
- Fourth position: 3 choices
- Fifth position: 2 choices
- Total = 6 × 5 × 4 × 3 × 2 = 720 numbers
Using formula: 6P5 = 6!/(6-5)! = 720/1 = 720
Special Case - Numbers starting with specific digit:
How many 4-digit even numbers can be formed from digits 1,2,3,4,5 without repetition?
- Last digit must be even → 2 choices (2 or 4)
- Case 1: Last digit = 2
- First digit: 4 choices (1,3,4,5)
- Second digit: 3 choices
- Third digit: 2 choices
- Total for case 1: 4×3×2 = 24
- Case 2: Last digit = 4 → Similar calculation: 24
- Total = 24 + 24 = 48 numbers
2. Alphabet-Based Permutation
Arranging letters of words with various conditions.
Basic Word Arrangement
Total arrangements = n! / (p₁! × p₂! × ...)
Where p₁, p₂ are counts of identical letters
Example: Arrange letters of \"MISSISSIPPI\"
Total letters = 11
M: 1, I: 4, S: 4, P: 2
Total arrangements = 11! / (4! × 4! × 2!) = 34,650 ways
Vowels Together
Treat all vowels as a single unit/block.
Example Problem: Arrange \"BHARAT\" with vowels always together
- Word: BHARAT → Letters: B,H,A,R,A,T (6 letters, 2 A\'s)
- Vowels: A,A → Treat as single block [AA]
- Now arrange: B, H, R, T, [AA] → 5 items to arrange
- Arrangements of 5 items = 5! = 120
- Vowels within block [AA] can arrange in 2!/2! = 1 way (both A\'s identical)
- Total arrangements = 5! × 1 = 120 ways
Example 2: Arrange \"EDUCATION\" with vowels together
- Word: EDUCATION → Vowels: E,U,A,I,O (5 distinct vowels)
- Consonants: D,C,T,N (4 consonants)
- Treat vowels as block [E U A I O] → 1 item
- Now arrange: D,C,T,N,[vowel block] → 5 items = 5! = 120
- Vowels within block can arrange in 5! = 120 ways
- Total = 5! × 5! = 120 × 120 = 14,400 ways
Consonants Together
Treat all consonants as a single unit/block.
Example Problem: Arrange \"AMISHA\" with consonants always together
- Word: AMISHA → Letters: A,M,I,S,H,A (6 letters, 2 A\'s)
- Consonants: M,S,H → Treat as single block [MSH]
- Now arrange: A, A, I, [MSH] → 4 items with 2 A\'s identical
- Arrangements of 4 items with repetition = 4!/2! = 12
- Consonants within block [MSH] can arrange in 3! = 6 ways
- Total arrangements = (4!/2!) × 3! = 12 × 6 = 72 ways
Vowels Never Together
Calculate total arrangements minus arrangements with vowels together.
Example: Arrange \"COMPUTER\" with no two vowels together
- Total letters: 8 (C,O,M,P,U,T,E,R)
- Vowels: O, U, E (3 vowels)
- Consonants: C, M, P, T, R (5 consonants)
- First arrange consonants: 5! = 120 ways
- Create slots for vowels: _C_M_P_T_R_ → 6 slots available
- Choose 3 slots from 6: ⁶C₃ = 20 ways
- Arrange 3 vowels in chosen slots: 3! = 6 ways
- Total = 120 × 20 × 6 = 14,400 ways
Rank of a Word
The position of a word when all possible arrangements are listed in alphabetical order.
Step-by-Step Method to Find Rank
- Arrange all letters of the word in alphabetical order
- For each position, count words starting with letters that come before the target word\'s letter
- Move to next position and repeat
- Add 1 for the target word itself
- Sum all counts to get the rank
Example Problem: Find rank of \"ROHIT\" in dictionary order
- Alphabetical order of letters: H, I, O, R, T
- Words starting with H: H _ _ _ _ → 4! = 24 words
- Words starting with I: I _ _ _ _ → 4! = 24 words
- Words starting with O: O _ _ _ _ → 4! = 24 words
- Words starting with R:
- RH _ _ _ → 3! = 6 words
- RI _ _ _ → 3! = 6 words
- RO:
- ROH _ _ → 2! = 2 words (ROHIT and ROH??)
- ROHIT is 1st word in this group
- Total count before ROHIT: 24 + 24 + 24 + 6 + 6 + (2-1) = 85
- Rank of ROHIT = 85 + 1 = 86th position
Example 2: Find rank of \"BLOCK\"
- Alphabetical order: B, C, K, L, O
- Words starting with B:
- BC _ _ _ → 3! = 6 words
- BK _ _ _ → 3! = 6 words
- BL:
- BLC _ _ → 2! = 2 words
- BLK _ _ → 2! = 2 words
- BLOCK is 1st in remaining
- Total count before BLOCK: 6 + 6 + 2 + 2 = 16
- Rank of BLOCK = 16 + 1 = 17th position
Quick Formula for Words with No Repeated Letters:
Rank = 1 + Σ (Number of letters lexicographically smaller × (n-1)!)
For \"ROHIT\":
R: 3 letters smaller (H,I,O) × 4! = 3×24 = 72
O: 2 letters smaller (H,I) × 3! = 2×6 = 12
H: 0 letters smaller × 2! = 0
I: 0 letters smaller × 1! = 0
T: No calculation needed
Rank = 72 + 12 + 1 = 85th (Note: This gives position before adding 1)
Circular Permutation
Definition and Formula
Arrangement of objects in a circle where rotations are considered the same arrangement.
When Clockwise & Anti-clockwise are Different
Where n = number of objects
Example Problem: How many ways can 5 people sit around a circular table?
Number of arrangements = (5 - 1)! = 4! = 4×3×2×1 = 24 ways
Explanation: Fix one person\'s position to eliminate identical rotations, arrange remaining 4.
When Clockwise & Anti-clockwise are Same
Where n = number of objects
Example Problem: How many different necklaces can be made with 10 distinct beads?
Number of arrangements = (10 - 1)! / 2 = 9! / 2 = 181,440 ways
Explanation: In necklaces, flipping gives same arrangement.
Special Cases
Circular Arrangement with Restrictions
Example: 5 couples (10 people) sit around a table with each couple sitting together
- Treat each couple as single unit → 5 units
- Arrange 5 units around table: (5-1)! = 4! = 24
- Each couple can switch places: 2 ways per couple
- Total = 24 × (2⁵) = 24 × 32 = 768 ways
Key People Not Sitting Together
Example: 8 people sit around table, 2 specific people not adjacent
- Total arrangements without restriction: (8-1)! = 7! = 5,040
- Treat the 2 people as single unit: 7 units → (7-1)! = 6! = 720
- The 2 people can arrange within unit: 2! = 2
- Arrangements with them together: 720 × 2 = 1,440
- Arrangements with them apart: 5,040 - 1,440 = 3,600 ways
Combination - Selection without Order
Definition and Formula
Combination refers to selecting objects without considering the order of selection. The order does NOT matter in combination.
Where:
n = Total number of objects
r = Number of objects to select
Key Properties:
- nC₀ = 1 (Selecting nothing)
- nC₁ = n (Selecting 1 from n)
- nCₙ = 1 (Selecting all)
- nCr = nC₍ₙ₋ᵣ₎ (Complementary combination)
- ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ (Pascal\'s Rule)
Types of Combination Problems
1. Selection from Multiple Groups
Example Problem: From 12 men and 15 women, select 10 men and 14 women for a committee. How many ways?
- Select 10 men from 12: ¹²C₁₀ = ¹²C₂ = (12×11)/(2×1) = 66
- Select 14 women from 15: ¹⁵C₁₄ = ¹⁵C₁ = 15
- Total ways = 66 × 15 = 990 ways
Example 2: Committee of 5 with at least 2 women from 6 men and 4 women
- Case 1: 2 women, 3 men: ⁴C₂ × ⁶C₃ = 6 × 20 = 120
- Case 2: 3 women, 2 men: ⁴C₃ × ⁶C₂ = 4 × 15 = 60
- Case 3: 4 women, 1 man: ⁴C₄ × ⁶C₁ = 1 × 6 = 6
- Total = 120 + 60 + 6 = 186 ways
2. Geometric Combinations
Example Problem: 9 non-collinear points on a plane. How many triangles can be formed?
Need 3 points to form triangle (order doesn\'t matter)
Number of triangles = ⁹C₃ = (9×8×7)/(3×2×1) = 84 triangles
Example 2: 12 vertical and 10 horizontal lines. How many rectangles?
Rectangle needs 2 vertical and 2 horizontal lines
- Choose 2 vertical from 12: ¹²C₂ = (12×11)/(2×1) = 66
- Choose 2 horizontal from 10: ¹⁰C₂ = (10×9)/(2×1) = 45
- Total rectangles = 66 × 45 = 2,970 rectangles
Example 3: How many diagonals in a polygon with n sides?
Total lines joining vertices = ⁿC₂
Subtract n sides (not diagonals)
Number of diagonals = ⁿC₂ - n = n(n-1)/2 - n = n(n-3)/2
For octagon (n=8): 8(8-3)/2 = 8×5/2 = 20 diagonals
3. Combination with Repetition/Selection
When items can be selected multiple times (like selecting fruits from a basket with unlimited quantity of each type).
ⁿ⁺ʳ⁻¹Cᵣ or (n + r - 1)! / [r! × (n - 1)!]
Example: Select 5 fruits from 3 types (apples, oranges, bananas) with unlimited supply
n = 3 types, r = 5 selections
Number of ways = ³⁺⁵⁻¹C₅ = ⁷C₅ = ⁷C₂ = (7×6)/(2×1) = 21 ways
Important Formulas and Relationships
nPr = nCr × r!
(Arrangement = Selection × arrangement of selected)
• Number of ways to divide n distinct items into groups of sizes a, b, c (a+b+c=n):
n! / (a! × b! × c!)
• When groups are identical: Divide by number of identical groups factorial
Number of handshakes among n people = ⁿC₂ = n(n-1)/2
Permutations & Combinations: Frequently Asked Questions
What\'s the main difference between permutation and combination?
Permutation considers the order/arrangement of items (ABC is different from BAC), while combination only considers selection (ABC is same as BAC when selecting 3 items).
When to use (n-1)! in circular permutation?
Use (n-1)! when arranging n distinct objects in a circle and clockwise/anti-clockwise are considered different. When they\'re considered same (like in necklaces), use (n-1)!/2.
How to handle repeated letters in word arrangement?
Divide total permutations (n!) by factorial of counts of each repeated letter. For \"BANANA\": 6!/(3!×2!) = 60 arrangements (3 A\'s, 2 N\'s).
What is the complement principle in combinations?
Instead of counting selections with conditions, count total selections minus selections without the condition. Often easier: \"at least 2\" = total - (0 or 1).
How to find rank of word with repeated letters?
For each position, count arrangements of letters before the target letter, adjusting for repetitions. Use factorial division for each counting step.
Practice Problems
Problem 1
How many 6-digit numbers can be formed using digits 0-9 if no digit repeats and number must be divisible by 10?
Problem 2
In how many ways can 5 Indians, 4 Americans, and 3 Russians sit in a row so that same nationality people sit together?
Problem 3
A committee of 7 with 4 men and 3 women is to be formed from 6 men and 5 women. In how many ways if 2 specific men refuse to serve together?
Problem 4
Find rank of \"MATHS\" when letters arranged alphabetically.
Frequently Asked Questions
What is Permutation and Combination Explained with Formulas and Examples?
Permutation and Combination Explained with Formulas and Examples is an important aptitude topic used in competitive exams that tests your logical reasoning and problem-solving abilities.
Is Permutation and Combination Explained with Formulas and Examples important for competitive exams?
Yes, Permutation and Combination Explained with Formulas and Examples is frequently asked in SSC, Bank, CAT, TCS, and other placement exams. It's essential to master this topic for better scores.
How to prepare Permutation and Combination Explained with Formulas and Examples easily?
Practice solved examples, learn formulas and shortcuts, and attempt practice questions regularly to master Permutation and Combination Explained with Formulas and Examples.
What are the important formulas in Permutation and Combination Explained with Formulas and Examples?
Key formulas vary by topic, but generally include basic concepts, shortcuts, and standard problem-solving approaches specific to Permutation and Combination Explained with Formulas and Examples.
How many questions come from Permutation and Combination Explained with Formulas and Examples?
Typically 5-10 questions come from Permutation and Combination Explained with Formulas and Examples in most competitive exams, making it a high-scoring section.
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