Permutation Combination FOR Placement

Meta Description: Master Permutation & Combination for Placement Exams 2026. Learn nPr/nCr formulas, circular arrangements, distribution tricks & solved PYQs from top companies.

Introduction

Permutation and Combination is a cornerstone of quantitative aptitude in campus recruitment drives. In 2025-2026 placement cycles, service-based giants like TCS, Wipro, HCL, Accenture, and Cognizant consistently feature 2 to 4 questions from this topic per aptitude section. The weightage typically ranges between 8% to 12% of the overall math/quant module. Beyond direct questions, mastery of arrangements and selections directly strengthens probability, logical reasoning, and data interpretation sections. Recruiters use these problems to evaluate a candidate’s analytical structuring ability, attention to constraints, and speed under pressure. This guide breaks down core concepts, company-tested problem patterns, and time-saving shortcuts to help you secure maximum accuracy.

Key Formulas & Concepts

Understanding when to apply each formula is more critical than memorization. Placement exams test conceptual clarity through constraints.

• Factorial (n!): Product of all positive integers ≤ n. n! = n × (n-1) × ... × 2 × 1. By definition, 0! = 1.
• Permutation (nPr): Used for arrangements where order matters.
  nPr = n! / (n - r)!
• Combination (nCr): Used for selections where order does not matter.
  nCr = n! / [r!(n - r)!]
• Circular Permutation: Arranging n distinct objects in a circle. Rotations are identical, so fix one position.
  Total = (n - 1)!
• Arrangements with Identical Objects: When n items contain duplicates (p1, p2, ... identical).
  Total = n! / (p1! × p2! × ...)
• Selections with Repetition Allowed: Choosing r items from n types, repetition allowed.
  Total = n+r-1 C r
• Distribution Basics:
  • Distinct items to distinct boxes (no restriction): n^r
  • Distinct items to distinct boxes (at least one per box): Use inclusion-exclusion or Stirling numbers of the second kind multiplied by r!
  • Identical items to distinct boxes (empty allowed): n+r-1 C r-1 (Stars & Bars)

Solved Examples (Basic Level)

Q1. How many 4-letter words can be formed from the letters of the word ‘LOGIC’ without repetition?
Solution: 5 distinct letters, select & arrange 4. 5P4 = 5! / (5-4)! = 120 / 1 = 120 words.

Q2. In a class of 12 students, how many ways can a committee of 3 be selected?
Solution: Order doesn’t matter. 12C3 = 12! / (3! × 9!) = (12×11×10) / (3×2×1) = 220 ways.

Q3. Find the value of 8P3 + 5C2.
Solution: 8P3 = 8×7×6 = 336. 5C2 = 10. Total = 336 + 10 = 346.

Q4. How many ways can 5 people sit around a circular table?
Solution: Circular arrangement = (5-1)! = 4! = 24 ways.

Q5. How many 3-digit numbers can be formed using digits 1, 2, 3, 4, 5 if repetition is allowed?
Solution: Each of the 3 positions has 5 choices. Total = 5 × 5 × 5 = 125 numbers.

Practice Questions (Medium Level)

Q6. In how many ways can the letters of ‘MATHEMATICS’ be arranged if vowels always stay together?
Solution: Vowels: A, E, A, I (4, with 2 A’s). Consonants: M, T, H, M, T, C, S (7, with 2 M’s, 2 T’s). Treat vowels as 1 block. Total units = 8. Arrangements = 8! / (2! × 2!) = 10080. Internal vowel arrangements = 4! / 2! = 12. Total = 10080 × 12 = 120960.

Q7. From 6 boys and 5 girls, a team of 4 must have at least 1 girl. Find the number of ways.
Solution: Total ways without restriction = 11C4 = 330. Ways with 0 girls (all boys) = 6C4 = 15. Valid = 330 - 15 = 315 ways.

Q8. How many ways can 8 distinct books be distributed among 3 students such that each gets at least one book?
Solution: Total unrestricted = 3^8 = 6561. Subtract cases where 1 or 2 students get 0. Using inclusion-exclusion: 3^8 - 3C1×2^8 + 3C2×1^8 = 6561 - 3×256 + 3×1 = 6561 - 768 + 3 = 5796.

Q9. Find the number of diagonals in a convex polygon of 10 sides.
Solution: Diagonals = nC2 - n = 10C2 - 10 = 45 - 10 = 35.

Q10. How many 5-digit numbers divisible by 5 can be formed using 0,1,2,3,4,5 without repetition?
Solution: Divisible by 5 ⇒ last digit is 0 or 5.
Case 1: Ends with 0. First digit ≠ 0. Remaining 4 positions from 1,2,3,4,5 ⇒ 5P4 = 120.
Case 2: Ends with 5. First digit can be 1,2,3,4 (not 0 or 5) ⇒ 4 choices. Next 3 positions from remaining 4 digits (including 0) ⇒ 4 × 4P3 = 4 × 24 = 96.
Total = 120 + 96 = 216.

Q11. In how many ways can 6 identical balls be placed in 4 distinct boxes if boxes can be empty?
Solution: Stars & Bars: n+r-1 C r-1 = 6+4-1 C 4-1 = 9C3 = 84.

Q12. How many words (with/without meaning) can be formed from ‘ENGINEER’?
Solution: 8 letters: E×3, N×2, G×1, I×1, R×1. Total = 8! / (3! × 2!) = 40320 / 12 = 3360.

Q13. A committee of 5 is to be formed from 7 men and 4 women with exactly 2 women.
Solution: Choose 2 women = 4C2 = 6. Choose 3 men = 7C3 = 35. Total = 6 × 35 = 210.

Advanced Questions

Q14. How many ways can 10 distinct chocolates be distributed among 3 distinct children such that one child gets 2, another gets 3, and the third gets 5?
Solution: First partition chocolates: 10! / (2! 3! 5!) = 3024. Since children are distinct and quotas are fixed, no extra permutation of quotas is needed. Total = 3024.

Q15. Find the number of ways to seat 7 people around a circular table if two specific persons must always sit opposite each other (assume even spacing isn’t required, just not adjacent).
Solution: Fix A at position 1. B must not sit adjacent to A. In a 7-seat circle, A has 2 adjacent seats. Remaining valid seats for B = 7 - 1 - 2 = 4. Remaining 5 people = 5!. Total = 4 × 120 = 480.

Q16. How many 4-digit numbers > 3000 can be formed from {1,2,3,4,5,6} without repetition?
Solution: First digit must be 3,4,5,6 (4 choices). Remaining 3 digits from remaining 5 = 5P3 = 60. Total = 4 × 60 = 240.

Q17. In how many ways can 5 boys and 5 girls stand in a row if they alternate?
Solution: Two patterns: BGBGBG... or GBGBGB... For each, arrange boys = 5!, girls = 5!. Total = 2 × 5! × 5! = 2 × 120 × 120 = 28,800.

Q18. Find the number of subsets of {1,2,3,4,5,6,7} that contain exactly 4 elements and at least one even number.
Solution: Total 4-element subsets = 7C4 = 35. Odd numbers = 1,3,5,7 (4 odds). Subsets with only odds = 4C4 = 1. Valid = 35 - 1 = 34.

Common Mistakes to Avoid

• Confusing order vs selection: Using nPr when the problem asks for groups/committees, or nCr when positions/seats matter.
• Ignoring identical items: Forgetting to divide by factorial of repeated letters or objects, leading to massive overcounting.
• Circular vs Linear trap: Applying n! for circular arrangements without subtracting 1, or forgetting the (n-1)!/2 rule for unoriented circles (like necklaces).
• Zero factorial error: Assuming 0! = 0 instead of 1, which breaks base-case calculations in combinations.
• Constraint blind spots: Missing hidden conditions like “first digit cannot be 0” or “at least one” requiring complementary counting.
• Repetition misapplication: Using n^r when items are identical, or n+r-1 C r when order matters.

Shortcut Tricks

1. Complementary Counting for “At Least One”
Instead of summing 1, 2, 3... cases, calculate Total - None.
Example: Q7 above used 11C4 - 6C4 instead of summing (1W+3M)+(2W+2M)+(3W+1M).

2. Fix & Multiply for Circular Arrangements
Always fix one person to remove rotational symmetry. Remaining = (n-1)!. If direction doesn’t matter (e.g., beads), divide by 2.
Example: 6 people around table = 5! = 120. Necklace = 120/2 = 60.

3. Grouping Method for “Always Together”
Treat the group as 1 super-unit. Arrange super-unit with others, then multiply by internal arrangements.
Example: Q6 vowels together = 8!/4! logic applied internally.

4. Stars & Bars for Identical Distribution
For distributing n identical items to r distinct boxes:
Ways = n+r-1 C r-1.
Example: 5 identical pens to 3 students = 5+3-1 C 3-1 = 7C2 = 21.

Previous Year Questions

Q19. (TCS NQT) How many ways can the digits 1, 2, 3, 4, 5, 6 be arranged so that 3 and 5 are never adjacent?
Solution: Total = 6! = 720. Treat 3&5 as one block = 5! × 2! = 240. Never adjacent = 720 - 240 = 480.

Q20. (Accenture) A password consists of 2 distinct letters followed by 3 distinct digits. How many passwords?
Solution: Letters: 26P2 = 26×25 = 650. Digits: 10P3 = 10×9×8 = 720. Total = 650 × 720 = 4,68,000.

Q21. (Wipro) In how many ways can 4 prizes be distributed among 3 students if a student can win multiple prizes?
Solution: Each prize has 3 choices. Independent distribution = 3^4 = 81.

Q22. (HCL) How many triangles can be formed by joining the vertices of a regular octagon?
Solution: 8 vertices, choose any 3 = 8C3 = 56. No three collinear in regular polygon, so all form triangles.

Q23. (Infosys) Number of ways to arrange ‘BANANA’ so vowels occupy only even positions.
Solution: Positions 1 2 3 4 5 6. Even = 2,4,6. Vowels: A,A,A (3). Must fill 3 even spots: 3!/3! = 1 way. Consonants B,N,N fill odd spots 1,3,5: 3!/2! = 3. Total = 1 × 3 = 3.

Quick Revision

• nPr = Order matters (seats, ranks, passwords)
• nCr = Order doesn’t matter (teams, subsets, committees)
• Circular = (n-1)! (divide by 2 for flip symmetry)
• Identical items = Divide by p! for each duplicate group
• “Always together” = Group as 1, then multiply internal arrangements
• “Never together” = Total – Together
• “At least one” = Total – None
• Identical distribution = n+r-1 C r-1 (Boxes can be empty)
• 0! = 1, 1! = 1, nCr = nC(n-r)
• Practice constraint mapping before plugging into formulas.