Permutation and Combination Solved Examples 2026 (40+ Problems)

Last Updated: June 2026

This page collects 40+ solved permutation and combination problems at placement-exam difficulty, organized from basics through arrangements, selections, circular cases, and constraints. Each problem has a full step-by-step solution. Candidates report this topic as a reliable scorer in TCS, Infosys, and Wipro quantitative sections once the permutation-versus-combination decision becomes automatic.

Formula Reference

Section A: Basics (Problems 1-10)

Problem 1. Evaluate 5!. Solution: 5 × 4 × 3 × 2 × 1 = 120.

Problem 2. Evaluate 6P2. Solution: 6!/(6-2)! = 6 × 5 = 30.

Problem 3. Evaluate 7C3. Solution: 7!/(3! × 4!) = (7 × 6 × 5)/(3 × 2 × 1) = 35.

Problem 4. Evaluate 8C8. Solution: 8!/(8! × 0!) = 1.

Problem 5. Evaluate 10C1. Solution: 10.

Problem 6. How many 3-digit numbers can be formed from digits 1 to 9 without repetition? Solution: 9 × 8 × 7 = 504.

Problem 7. How many ways to arrange 4 distinct books on a shelf? Solution: 4! = 24.

Problem 8. In how many ways can 3 prizes be given to 5 students if no student gets more than one? Solution: 5P3 = 5 × 4 × 3 = 60.

Problem 9. How many ways to choose 2 fruits from 6 different fruits? Solution: 6C2 = 15.

Problem 10. Evaluate 9P0. Solution: nP0 = 1.

Section B: Word Arrangements (Problems 11-20)

Problem 11. How many arrangements of the letters of the word LEADER? Solution: 6 letters with E repeated twice: 6!/2! = 720/2 = 360.

Problem 12. How many arrangements of the word BANANA? Solution: 6 letters, A repeated 3 times, N repeated 2 times: 6!/(3! × 2!) = 720/12 = 60.

Problem 13. How many arrangements of the word APPLE? Solution: 5 letters, P repeated twice: 5!/2! = 60.

Problem 14. In how many arrangements of MOTHER do the vowels come together? Solution: Vowels O, E treated as one block: 5 units arranged in 5! ways, vowels among themselves in 2! ways: 120 × 2 = 240.

Problem 15. How many arrangements of the word SUCCESS? Solution: 7 letters, S repeated 3 times, C repeated twice: 7!/(3! × 2!) = 5040/12 = 420.

Problem 16. In how many arrangements of GARDEN do the two vowels never come together? Solution: Total = 6! = 720. Vowels together = 5! × 2! = 240. Never together = 720 - 240 = 480.

Problem 17. How many 4-letter words from the letters of EQUATION using each once? Solution: 8 distinct letters, choose and arrange 4: 8P4 = 8 × 7 × 6 × 5 = 1680.

Problem 18. How many arrangements of MISSISSIPPI? Solution: 11 letters: M once, I four times, S four times, P twice: 11!/(4! × 4! × 2!) = 39916800/1152 = 34650.

Problem 19. In how many arrangements of the word LOGIC do all consonants come together? Solution: Consonants L, G, C as one block plus vowels O, I: 3 units arranged in 3! ways, consonants among themselves in 3! ways: 6 × 6 = 36.

Problem 20. How many arrangements of the word BOOK? Solution: 4 letters, O repeated twice: 4!/2! = 12.

Section C: Selections and Committees (Problems 21-32)

Problem 21. From 8 men, choose a committee of 3. Solution: 8C3 = 56.

Problem 22. From 6 men and 4 women, choose a committee of 3 with exactly 2 men. Solution: 6C2 × 4C1 = 15 × 4 = 60.

Problem 23. From 5 boys and 4 girls, form a team of 4 with at least one girl. Solution: Total 9C4 = 126, minus all-boys 5C4 = 5, so 126 - 5 = 121.

Problem 24. In how many ways can 11 players be chosen from 15? Solution: 15C11 = 15C4 = 1365.

Problem 25. From 7 questions, a student must answer 5. In how many ways? Solution: 7C5 = 21.

Problem 26. A bag has 4 red and 3 blue balls. Choose 3 balls. In how many ways are exactly 2 red? Solution: 4C2 × 3C1 = 6 × 3 = 18.

Problem 27. From 10 points, no three collinear, how many triangles? Solution: 10C3 = 120.

Problem 28. How many diagonals does a hexagon have? Solution: 6C2 - 6 = 15 - 6 = 9.

Problem 29. From a deck of 52, in how many ways can 2 kings be chosen? Solution: 4C2 = 6.

Problem 30. A committee of 5 from 9 people must include a particular person. In how many ways? Solution: Fix that person, choose 4 from remaining 8: 8C4 = 70.

Problem 31. From 6 different books, in how many ways can a student select at least one? Solution: 2^6 - 1 = 64 - 1 = 63.

Problem 32. How many handshakes occur if 12 people each shake hands once with every other? Solution: 12C2 = 66.

Section D: Circular and Constraints (Problems 33-42)

Problem 33. In how many ways can 6 people sit around a round table? Solution: (6-1)! = 5! = 120.

Problem 34. In how many ways can 5 beads form a necklace? Solution: (5-1)!/2 = 24/2 = 12, since a necklace can be flipped.

Problem 35. In how many ways can 7 people sit around a table if 2 particular people sit together? Solution: Treat the pair as one unit: (6-1)! × 2! = 120 × 2 = 240.

Problem 36. In how many ways can 4 boys and 4 girls sit around a table alternately? Solution: Fix boys in (4-1)! = 6 ways, then girls in the 4 gaps in 4! = 24 ways: 6 × 24 = 144.

Problem 37. How many 4-digit even numbers can be formed from 1, 2, 3, 4, 5 without repetition? Solution: Last digit must be 2 or 4, so 2 choices; remaining 3 places from 4 digits: 2 × 4 × 3 × 2 = 48.

Problem 38. How many 3-digit numbers greater than 600 from digits 4, 5, 6, 7, 8 without repetition? Solution: First digit must be 6, 7, or 8, so 3 choices; remaining 2 places from 4 digits: 3 × 4 × 3 = 36.

Problem 39. In how many ways can the letters of TABLE be arranged so that vowels occupy the end positions? Solution: Vowels A, E in the 2 end positions in 2! ways, consonants T, B, L in the 3 middle positions in 3! ways: 2 × 6 = 12.

Problem 40. How many 5-digit numbers can be formed from 0, 1, 2, 3, 4 without repetition (no leading zero)? Solution: Total arrangements 5! = 120, minus those with leading zero 4! = 24, so 120 - 24 = 96.

Problem 41. In how many ways can 3 men and 2 women stand in a row so that women are never adjacent? Solution: Arrange 3 men in 3! = 6 ways, creating 4 gaps; place 2 women in 4 gaps in 4P2 = 12 ways: 6 × 12 = 72.

Problem 42. From the word NUMBER, how many arrangements start with a vowel? Solution: Vowels are U and E, so 2 choices for the first letter, remaining 5 letters in 5! = 120 ways: 2 × 120 = 240.

The Decision That Decides Everything: Order Or Not

The single most important judgment in this topic is whether order matters. Everything else follows from that call. Ask yourself: if I rearrange the same chosen items, is that a different outcome? If yes, order matters and you use permutations. If no, order does not matter and you use combinations.

Arranging three people in three chairs is a permutation because ABC and ACB are different seatings. Choosing three people for a committee is a combination because the committee ABC is the same regardless of the order you named them. Forming a three-digit number is a permutation because 123 and 321 are different numbers. Selecting three cards for a hand is a combination because the hand is the same set whatever order they were dealt. Candidates report that getting this one judgment right resolves most of the difficulty, because once you know which tool applies, the formula is mechanical.

Handling The Common Constraints

Beyond the basic order judgment, a few constraint patterns recur and each has a standard handling.

When certain items must stay together, treat the group as a single unit, arrange the units, then multiply by the internal arrangements of the group. When items must never be together, compute the total arrangements and subtract the together case, or use the gaps method by arranging the others first and slotting the restricted items into the gaps between them. When letters repeat, divide the total factorial by the factorial of each repeat count to remove duplicate orderings. When the problem says at least one, compute the total and subtract the none case rather than summing the messy at-least-one cases. When arranging in a circle, fix one position to remove rotational duplicates, giving (n minus one) factorial, and divide by two further if reflections count as the same. These five constraint patterns cover the large majority of placement permutation and combination questions.

Key Takeaways

Candidates report that the decisive skill in this topic is recognizing whether order matters before reaching for a formula. Once the permutation-versus-combination call is automatic, the rest is arithmetic with factorials.

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Frequently Asked Questions

When do I use permutation versus combination?

Use permutation when order matters, such as arranging people in a row or forming numbers. Use combination when order does not matter, such as selecting a team or a committee. Candidates report that asking whether rearranging the same items counts as a new outcome is the quickest way to decide which applies.

What is the formula for circular arrangements?

For n distinct items in a circle, the number of arrangements is (n minus 1) factorial, because rotations of the same arrangement are counted once. If clockwise and anticlockwise are considered the same, as with a garland, divide by two to get (n minus 1) factorial over two.

How do I handle arrangements with repeated letters?

Divide the total factorial by the factorial of each repeated letter's count. For a word with n letters where one letter repeats p times and another q times, the arrangements equal n factorial divided by (p factorial times q factorial). This removes the duplicate orderings of identical letters.