How to Solve Permutation and Combination Problems

How to Solve Permutation and Combination Problems

(For SSC, Bank, Railway Exams – Ace Your Exam with Confidence!)

Introduction

"Mastering permutations and combinations can add 5–10 marks to your SSC/Bank/Railway exam—enough to push you from ‘just passing’ to ‘top ranker’! These questions appear in every competitive exam, and if you follow this exact method, you’ll solve them in under 60 seconds—no guessing, no panic."

What You Need To Know First

Before diving in, ensure you understand:
1. Factorials (n!) – The product of all positive integers up to n (e.g., 4! = 4 × 3 × 2 × 1 = 24).
2. Difference between "order matters" and "order doesn’t matter" – This decides whether you use permutation (P) or combination (C).
3. Basic counting principle – If Event A has m ways and Event B has n ways, total ways = m × n.

Key Vocabulary

Formulas To Know

1. Permutation (Order Matters)

Formula: [ P(n, r) = \frac{n!}{(n - r)!} ] - n = Total items - r = Items to arrange - MEMORISE THIS (Not always given on exam sheet)

When to use: Arranging people in a line, forming passwords, ranking teams.

2. Combination (Order Doesn’t Matter)

Formula: [ C(n, r) = \frac{n!}{r!(n - r)!} ] - n = Total items - r = Items to select - MEMORISE THIS (Often given, but know it anyway)

When to use: Selecting a team, choosing lottery numbers, forming groups.

3. Permutation with Repetition

Formula: [ n^r ] - n = Choices per position - r = Number of positions - MEMORISE THIS

When to use: Forming 3-digit numbers with digits 0–9 (repetition allowed).

4. Permutation of Non-Distinct Items

Formula: [ \frac{n!}{p! \times q! \times \dots} ] - n = Total items - p, q = Number of identical items - MEMORISE THIS

When to use: Arranging letters in "MISSISSIPPI" (repeated letters).

Step-by-Step Method

Step 1: Read the Question Carefully

• Underline keywords: "arrange," "select," "order," "group," "password," "team."
• Ask: Does order matter? (If yes → Permutation. If no → Combination.)

Step 2: Identify Total Items (n) and Items to Choose (r)

• Count the total number of items (n).
• Count how many are being arranged/selected (r).

Step 3: Check for Repetition

• Can items be reused? (e.g., digits in a number, letters in a password)
• Yes → Use ( n^r )
• No → Use P(n, r) or C(n, r)

Step 4: Check for Identical Items

• Are there repeated items? (e.g., letters in "BOOK")
• Yes → Use non-distinct permutation formula
• No → Use standard P(n, r) or C(n, r)

Step 5: Apply the Correct Formula

• Plug n and r into the formula.
• Simplify factorials (cancel out common terms).

Step 6: Verify with a Quick Check

• Does the answer make sense? (e.g., P(5,2) = 20, not 100)
• If stuck, try a smaller example (e.g., P(3,2) = 6).

Worked Examples

Example 1 – Basic (Permutation)

Question: In how many ways can 5 students be seated in 3 chairs?

Solution:
1. Order matters? Yes (seating arrangement).
2. n = 5 (students), r = 3 (chairs).
3. Repetition? No (one student per chair).
4. Formula: ( P(5, 3) = \frac{5!}{(5-3)!} = \frac{5!}{2!} )
5. Calculate: ( 5! = 120 ) ( 2! = 2 ) ( \frac{120}{2} = 60 )

Answer: 60 ways.

What we did and why: - Since order matters (who sits where), we used permutation. - We canceled ( 2! ) to simplify calculation.

Example 2 – Medium (Combination with Restriction)

Question: A team of 4 is to be selected from 6 boys and 4 girls. In how many ways can the team have exactly 2 girls?

Solution:
1. Order doesn’t matter (team selection).
2. Total girls = 4, need 2 girls → ( C(4, 2) )
3. Total boys = 6, need 2 boys → ( C(6, 2) )
4. Use multiplication principle: ( C(4, 2) \times C(6, 2) )
5. Calculate: ( C(4, 2) = \frac{4!}{2!2!} = 6 ) ( C(6, 2) = \frac{6!}{2!4!} = 15 ) ( 6 \times 15 = 90 )

Answer: 90 ways.

What we did and why: - Since order doesn’t matter, we used combination. - We broke the problem into two parts (girls + boys) and multiplied.

Example 3 – Exam-Style (Permutation with Repetition)

Question: How many 3-digit numbers can be formed using digits 1, 2, 3, 4 if repetition is allowed?

Solution:
1. Order matters? Yes (123 ≠ 321).
2. Repetition allowed? Yes (e.g., 111, 122).
3. n = 4 (digits), r = 3 (positions).
4. Formula: ( n^r = 4^3 )
5. Calculate: ( 4 \times 4 \times 4 = 64 )

Answer: 64 numbers.

What we did and why: - Since digits can repeat, we used ( n^r ). - Each digit has 4 choices, and there are 3 positions.

Common Mistakes

Exam Traps

1-Minute Recap (Night Before Exam)

"Listen up—this is all you need to remember for permutations and combinations:

1. Order matters? → Permutation (P). Order doesn’t matter? → Combination (C).
2. Formulas:
3. P(n,r) = ( \frac{n!}{(n-r)!} ) (arranging)
4. C(n,r) = ( \frac{n!}{r!(n-r)!} ) (selecting)
5. Repetition allowed? → ( n^r )
6. Watch for traps:
7. "At least" = break into cases.
8. Repeated letters? → Divide by factorial of repeats.
9. Zero in digits? → Subtract invalid numbers.
10. Always simplify factorials first—cancel before multiplying.
11. Double-check: Does your answer make sense? (e.g., P(3,2) = 6, not 100).

You’ve got this—go solve those 5–10 marks like a pro!