Permutation and Combination

Permutation and Combination

Overview

Permutation and Combination forms the foundation of counting problems in IBPS Clerk Prelims. The marks here are essentially "free" for well-prepared candidates.

The fundamental distinction you must master: Permutation counts arrangements where order matters (who sits where), while Combination counts selections where order doesn't matter (who is chosen). IBPS Clerk focuses on basic applications—forming numbers, selecting teams, arranging people in rows—rather than complex multi-stage problems seen in PO or CAT exams.

Mastering this topic also builds your foundation for Probability, as most probability problems require counting favorable and total outcomes using these very principles.

Key Concepts

• Fundamental Counting Principle: If one task can be done in m ways and another in n ways, both tasks together can be done in m × n ways. This is the backbone of all counting problems.
• Factorial (n!): The product of all positive integers from 1 to n. Example: 5! = 5 × 4 × 3 × 2 × 1 = 120. Remember: 0! = 1 and 1! = 1.
• Permutation (ⁿPᵣ): Number of ways to arrange r items from n distinct items where order matters. Use when the problem asks about "arrangements," "sequences," or "positions."
• Combination (ⁿCᵣ): Number of ways to select r items from n distinct items where order doesn't matter. Use when the problem asks about "selecting," "choosing," or "forming groups."
• Permutation with Repetition: When items can repeat, n choices for each of r positions gives nʳ arrangements.
• Combination with Constraints: Problems often add conditions like "at least," "at most," or "exactly"—break these into cases and add/subtract accordingly.
• Symmetry Property: ⁿCᵣ = ⁿCₙ₋ᵣ. Choosing 3 items from 10 equals choosing 7 items to leave behind.

Formulas / Key Facts

Must-Remember Factorials: 2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720, 7! = 5040

Worked Examples

Example 1: Basic Permutation

Problem: In how many ways can 5 students be arranged in a row for a photograph?

Solution:

• All 5 students are being arranged, and order matters (positions are distinct)
• This is arranging 5 items in 5 positions = 5!
• 5! = 5 × 4 × 3 × 2 × 1 = 120 ways

Example 2: Basic Combination

Problem: From a group of 8 employees, in how many ways can a committee of 3 be formed?

Solution:

• We are selecting 3 from 8, order doesn't matter (committee members have no ranks)
• This is ⁸C₃
• ⁸C₃ = 8!/(3! × 5!) = (8 × 7 × 6)/(3 × 2 × 1) = 336/6 = 56 ways

Example 3: Forming Numbers with Conditions

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

Solution:

• Hundreds place: 5 choices (any of the 5 digits)
• Tens place: 4 choices (one digit already used)
• Units place: 3 choices (two digits already used)
• Total = 5 × 4 × 3 = 60 numbers

Example 4: Selection with Constraints

Problem: A team of 4 is to be selected from 5 men and 4 women such that at least 2 women are included. How many ways?

Solution:

• Case 1: Exactly 2 women and 2 men = ⁴C₂ × ⁵C₂ = 6 × 10 = 60
• Case 2: Exactly 3 women and 1 man = ⁴C₃ × ⁵C₁ = 4 × 5 = 20
• Case 3: Exactly 4 women and 0 men = ⁴C₄ × ⁵C₀ = 1 × 1 = 1
• Total = 60 + 20 + 1 = 81 ways

Common Mistakes

• Confusing when order matters: Students use permutation for committee selection (where order doesn't matter) → Fix: Ask yourself, "Does swapping two selected items create a different outcome?" If no, use combination.
• Forgetting constraints on first digit: When forming numbers, the first digit cannot be 0 → Fix: Always fill the most restricted position first (hundreds place), then count remaining positions.
• Calculation errors with factorials: Computing 7!/(4! × 3!) by expanding everything → Fix: Cancel common terms immediately. 7!/(4! × 3!) = (7 × 6 × 5)/(3 × 2 × 1).
• Missing cases in "at least" problems: Calculating only one case when multiple satisfy the condition → Fix: List all valid cases systematically (at least 2 = exactly 2 + exactly 3 + exactly 4...).
• Double-counting in arrangements: Treating identical items as distinct when they're not → Fix: For words like COMMITTEE with repeated letters, divide by the factorial of repeated letter counts.

Quick Reference

• Order matters → Permutation (ⁿPᵣ) | Order doesn't matter → Combination (ⁿCᵣ)
• ⁿPᵣ = n!/(n-r)! and ⁿCᵣ = n!/[r!(n-r)!]
• Fundamental Principle: Multiply choices at each independent step
• "At least" problems: Break into cases or use Total − Unwanted cases
• First digit ≠ 0: Always handle the hundreds place separately when forming numbers
• ⁿCᵣ = ⁿCₙ₋ᵣ: Use the smaller value of r for faster calculation (¹⁰C₈ = ¹⁰C₂)