Permutation and Combination

Arrangements and selections

128 practice questionsPage updated 5 Apr 2026Review metadata pendingPractice questions

Permutation and Combination are fundamental concepts in counting problems. Permutation deals with arrangements where order matters, while Combination deals with selections where order doesn't matter. These concepts are essential for solving problems involving arrangements, selections, and probability calculations.

Key formulas and rules

nPr = n! / (n-r)! (Permutation of n items taken r at a time)

nCr = n! / [r! x (n-r)!] (Combination of n items taken r at a time)

nCr = nC(n-r) (Symmetry property)

nPr = nCr x r! (Relationship between permutation and combination)

nC0 + nC1 + nC2 + ... + nCn = 2^n (Sum of all combinations)

Circular permutation of n distinct objects = (n-1)!

Circular permutation when clockwise and anticlockwise are same = (n-1)!/2

Number of ways to arrange n items with p identical of one type, q of another = n! / (p! x q!)

nCr = n-1Cr-1 + n-1Cr (Pascal's Identity)

Key concepts

Fundamental Principle of Counting

Multiplication Principle: If one operation can be performed in 'm' ways and another in 'n' ways independently, then both operations together can be performed in m x n ways. Addition Principle: If two operations are mutually exclusive (either one OR the other), the total ways = m + n ways. Always determine whether to multiply (AND) or add (OR) based on the problem structure.

Permutations - Order Matters

Use permutations when the arrangement or order of selection is important. For example: forming numbers, arranging people in a line, assigning distinct positions, creating codes with different arrangements. Key scenarios: linear arrangements (nPr), circular arrangements ((n-1)!), arrangements with repetition (n^r), and arrangements with identical items (accounting for duplicates by dividing by factorial of identical counts).

Worked examples

Example 1

In how many ways can 5 people be seated around a circular table?

Solution

Step 1: Identify this as a circular permutation problem.
Step 2: For circular arrangements, rotations are considered the same arrangement.
Step 3: Fix one person's position to break the rotational symmetry.
Step 4: Arrange the remaining 4 people in 4! = 24 ways.
Step 5: Alternatively, use formula (n-1)! = (5-1)! = 4! = 24.
Answer: 24 ways

Example 2

A committee of 4 members is to be formed from 8 men and 6 women. In how many ways can this be done if the committee must have at least 2 women?

Solution

Step 1: Calculate total people = 8 + 6 = 14.
Step 2: 'At least 2 women' means 2 women, 3 women, or 4 women.
Step 3: Case 1 (2W, 2M): 6C2 x 8C2 = 15 x 28 = 420
Step 4: Case 2 (3W, 1M): 6C3 x 8C1 = 20 x 8 = 160
Step 5: Case 3 (4W, 0M): 6C4 x 8C0 = 15 x 1 = 15
Step 6: Total = 420 + 160 + 15 = 595
Answer: 595 ways

Example 3

How many 4-letter words can be formed from the letters of 'BANANA'?

Solution

Step 1: Letters in BANANA: B(1), A(3), N(2) - total 6 letters.
Step 2: We need 4-letter words. Cases based on letter repetition:
Step 3: Case 1 - All 4 letters different: Choose from {B, A, N} = only 3 distinct, need 4 different - impossible (0 ways).
Step 4: Case 2 - One letter repeats twice: Choose which letter repeats (A or N), choose 2 more from remaining.
Step 5: For AA: select 2 from {B, N} in 2C2 = 1 way, arrange 4!/(2!) = 12 ways. Total = 1 x 12 = 12
Step 6: For NN: select 2 from {B, A} in 2C2 = 1 way, arrange 4!/(2!) = 12 ways. Total = 1 x 12 = 12
Step 7: Case 3 - One letter repeats thrice (AAA): Select 1 from {B, N} in 2C1 = 2 ways, arrange 4!/(3!) = 4 ways. Total = 2 x 4 = 8
Step 8: Total words = 12 + 12 + 8 = 32
Answer: 32 words

Example 4

In how many ways can 6 identical balls be distributed into 4 distinct boxes such that no box is empty?

Solution

Step 1: This is distributing identical items into distinct boxes with constraint.
Step 2: Since no box can be empty, give 1 ball to each box first.
Step 3: Remaining balls = 6 - 4 = 2 balls to distribute freely.
Step 4: Use stars and bars: distribute 2 identical balls into 4 distinct boxes.
Step 5: Formula: (n+r-1)C(r-1) where n=balls, r=boxes.
Step 6: (2+4-1)C(4-1) = 5C3 = 10
Alternative: List partitions of 6 into 4 positive parts: (3,1,1,1), (2,2,1,1) and their permutations.
Step 7: For (3,1,1,1): 4 ways (choose which box gets 3).
Step 8: For (2,2,1,1): 4C2 = 6 ways (choose which 2 boxes get 2 each).
Step 9: Total = 4 + 6 = 10
Answer: 10 ways

Permutation and Combination

Key formulas and rules

Key concepts

Fundamental Principle of Counting

Permutations - Order Matters

Combinations - Order Doesn't Matter

Special Cases and Constraints

Circular and Restricted Arrangements

Grouping and Distribution Problems

Worked examples

Example 1

Example 2

Example 3

Example 4

Representative solved questions

Question 1

Question 2

Question 3

Common mistakes and useful habits

Ready to test your understanding?

Key formulas and rules

Key concepts

Fundamental Principle of Counting

Permutations - Order Matters

Combinations - Order Doesn't Matter

Special Cases and Constraints

Circular and Restricted Arrangements

Grouping and Distribution Problems

Worked examples

Example 1

Example 2

Example 3

Example 4

Representative solved questions

Question 1

Question 2

Question 3

Common mistakes and useful habits

Ready to test your understanding?

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