How to Solve: Permutations and Combinations (IIT JEE Guide)

How to Solve: Permutations and Combinations (IIT JEE Guide)

Hook: Mastering permutations and combinations unlocks 10-15% of IIT JEE Maths marks—from seating arrangements to probability, and even advanced topics like derangements. One wrong formula, and you lose 4-6 marks in seconds. This guide ensures you never mix them up again.

What You Need to Know First

1. Factorials – What is n! and how to simplify expressions like 5! / 3!.
2. Basic Counting – The difference between "AND" (multiply) and "OR" (add).
3. Set Theory – Understanding subsets, unions, and intersections.

Key Vocabulary

Formulas to Know

Step-by-Step Method

Step 1: Read the Problem Carefully

• Underline keywords: "arrange" (permutation), "select" (combination), "no two same" (no repetition), "order matters" (permutation).
• Circle numbers: n (total items), r (items to choose/arrange).

Step 2: Decide if Order Matters

• Order matters? → Permutation (P(n, r))
• Example: Arranging books on a shelf, forming passwords, ranking teams.
• Order doesn’t matter? → Combination (C(n, r))
• Example: Selecting a team, choosing lottery numbers, forming committees.

Step 3: Check for Repetition

• Repetition allowed? → Use nʳ (permutation) or C(n + r - 1, r) (combination).
• Example: Forming 3-digit numbers with digits 0-9 (repetition allowed).
• No repetition? → Use P(n, r) or C(n, r).
• Example: Arranging 3 distinct books from 5.

Step 4: Apply the Correct Formula

• Plug numbers into the formula.
• Simplify factorials (e.g., 5! / 3! = 5 × 4).

Step 5: Verify Constraints

• Are there restrictions? (e.g., "A must sit in the first seat," "no two identical items together").
• Adjust the formula (e.g., fix one item and arrange the rest).

Step 6: Calculate and Check Units

• Ensure the answer is an integer (no fractions or decimals).
• If the answer is too large, recheck for repetition or order.

Worked Example (Using Steps Above)

Problem: In how many ways can 5 distinct books be arranged on a shelf if 2 specific books must always be together?

Step 1: Read the Problem

• Keywords: "arranged" (order matters), "must always be together" (constraint).
• Numbers: n = 5 (total books), r = 5 (arranging all).

Step 2: Order Matters?

• Yes → Permutation.

Step 3: Repetition?

• No → Use P(n, r).

Step 4: Apply Formula

• Normally, P(5, 5) = 5! = 120.
• But 2 books must be together → Treat them as 1 unit.
• Now, we have 4 units (the pair + 3 other books).
• Ways to arrange 4 units: 4! = 24.
• Ways to arrange the 2 books within the pair: 2! = 2.
• Total ways = 4! × 2! = 24 × 2 = 48.

Step 5: Verify Constraints

• Constraint satisfied: The 2 books are always together.

Step 6: Check Units

• 48 is an integer → Valid.

Answer: 48 ways

What we did and why: We treated the 2 books as a single unit to satisfy the constraint, then multiplied by the internal arrangements of the pair. This is a common trick for "must be together" problems.

Worked Examples

Example 1 – Basic (Combination)

Problem: In how many ways can 3 students be selected from a class of 10?

Solution: 1. Order doesn’t matter → Combination. 2. No repetition → C(n, r). 3. C(10, 3) = 10! / (3! × 7!) = (10 × 9 × 8) / (3 × 2 × 1) = 120.

Answer: 120 ways

What we did and why: We used C(n, r) because the order of selection doesn’t matter (e.g., selecting Alice, Bob, Charlie is the same as Bob, Alice, Charlie).

Example 2 – Medium (Permutation with Constraint)

Problem: How many 4-digit numbers can be formed using digits 1-7 if no digit repeats and the number must be even?

Solution: 1. Order matters → Permutation. 2. No repetition → P(n, r). 3. Constraint: Number must be even → Last digit must be even (2, 4, 6).
- Step 1: Choose last digit (3 choices: 2, 4, 6).
- Step 2: Choose first 3 digits from remaining 6 digits: P(6, 3) = 6 × 5 × 4 = 120.
- Total ways = 3 × 120 = 360.

Answer: 360 numbers

What we did and why: We fixed the last digit first (to satisfy the "even" condition), then arranged the remaining digits. This is a constraint-first approach.

Example 3 – Exam-Style (Derangement)

Problem: In how many ways can 4 letters be placed into 4 envelopes such that no letter goes into its correct envelope?

Solution: 1. This is a derangement problem (no item in its original position). 2. Use derangement formula:
- D₄ = 9 (memorise small derangements: D₁ = 0, D₂ = 1, D₃ = 2, D₄ = 9).
- Alternatively, use the formula:
D₄ = 4! [1/0! - 1/1! + 1/2! - 1/3! + 1/4!] = 24 [1 - 1 + 0.5 - 0.1667 + 0.0417] ≈ 9.

Answer: 9 ways

What we did and why: We recognised this as a derangement (a common IIT JEE trap) and applied the formula directly. Always check for "no correct position" problems!

Common Mistakes

Exam Traps

1-Minute Recap (Night Before Exam)

"Listen up—this is your 60-second crash course for permutations and combinations. First, order matters? Permutation. Order doesn’t matter? Combination. Repetition allowed? Use nʳ or C(n + r - 1, r). No repetition? P(n, r) or C(n, r). Circular? (n - 1)!. Derangement? No item in its original position—use the formula or memorise small values. Constraints? Fix them first, then arrange the rest. And always check if the problem is tricking you with "at least one" or "exactly one." Now go crush that exam!