IPMAT Permutation and Combination Questions with Answers (2027

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Author : Lalita Vishwakarma

August 13, 2026

Overview: Do you want to master permutations and combinations for the IPMAT? This guide explains the core concepts, gives you the exact formulas, and then walks you through a large, fully verified set of IPMAT Permutation and Combination Practice Questions so you can solve them accurately and quickly in the IPMAT exam 2027.

What are the Permutation and combination questions in the IPMAT exam?

Direct answer

Permutation and combination questions test your ability to count possibilities efficiently using logical reasoning.

These problems ask you either to arrange objects (permutation) or to select objects (combination) under specific conditions.

Because the question types repeat almost every year, this is one of the highest-return topics in the IPMAT exam Quantitative Ability section — a few reliable marks here are among the easiest to secure.

Most Important IPMAT Permutation and Combination Practice Questions for 2027

Every answer in this section has been re-verified for accuracy. The questions are ordered from easy to hard so you can build up gradually. If your basics feel shaky, revisit the IPMAT maths syllabus before attempting them, then treat these as your first timed drill.

Q1. How many numbers between 100 and 1000 have at least one digit equal to 5?
(a) 215

(b) 225

(d) 252

Correct Answer:

(d) 252. Total 3-digit numbers = 900; those with no 5 = 8×9×9 = 648; so 900 − 648 = 252.

Q2. For five true/false questions, no student has all answers correct and no two students give the same sequence. What is the maximum number of students possible?
(a) 15

(b) 31

(d) 63

Correct Answer:

(b) 31. Total sequences = 2⁵ = 32; excluding the all-correct one leaves 31.

Q3. A mint prepares one calendar plate per February. How many distinct February calendars cover all future possibilities?
(a) 7

(b) 21

(d) 49

Correct Answer:

(c) 14. February is either 28 or 29 days (2 options) and can start on any of 7 weekdays, so 2×7 = 14.

Q4. Six MCQs: the first three have 4 choices each and the next three have 5 choices each. How many answer sequences are possible?

(a) 1000

(b) 8000

(d) 4000

Correct Answer:

(b) 8000. 4×4×4×5×5×5 = 64×125 = 8000.

Q5. In how many ways can 6 letters be posted in 5 letter boxes?

(a) 6!

(b) 6⁵

(d) 5⁶

Correct Answer:

d) 5⁶= 15625.Each of the 6 letters independently has 5 box choices.

Q6. There are 5 letters and 5 correctly addressed envelopes. In how many ways can the letters be placed so that not all land in the correct envelope?

(a) 32

(b) 31

(d) 120

Correct Answer

(c) 119. Total arrangements = 5! = 120; exactly one is all-correct, so 120 − 1 = 119.

Q7. 12 villages are split into 3 zones of 4. Villages in the same zone get 3 direct lines each pair; villages in different zones get 2 direct lines each pair. How many direct lines in total?

(a) 112

(b) 120

(d) 132

Correct Answer:

(c) 150. Same-zone pairs = 3×C(4,2) = 18 → 54 lines; different-zone pairs = 66 − 18 = 48 → 96 lines; total 150.

Q8. How many 3-digit numbers divisible by 3 can be formed using digits 2, 3, 4, 5 with repetition allowed?
(a) 22

(b) 6

(d) 21
Correct Answer:

(a) 22. Count 3-digit strings from {2,3,4,5} whose digit sum is a multiple of 3.

Q9. How many 4-digit numbers divisible by 4 can be formed using 0–6, first with repetition and then without repetition?
(a) 588, 208

(b) 720, 240

(d) 320, 124
Correct Answer:

(a) 588, 208. With repetition there are 588 such numbers; without repetition there are 208.

Q10. How many 4-digit even numbers can be formed using 1, 2, 3, 4, 5, 6 without repetition?
(a) 72

(b) 120

(d) 240
Correct Answer:

(c) 180. Even last digit → 3 choices; remaining three places = 5×4×3 = 60; total 3×60 = 180.

Q11. A 5-member committee is formed from 7 boys and 6 girls with at least 2 girls. In how many ways?
(a) 1056

(b) 462

(d) 252
Correct Answer:

(a) 1056. C(6,2)C(7,3) + C(6,3)C(7,2) + C(6,4)C(7,1) + C(6,5)C(7,0) = 525 + 420 + 105 + 6 = 1056.

Q12. How many 5-digit numbers using 1, 2, 3, 4, 5 without repetition are greater than 23,000?
(a) 48

(b) 60

(d) 120
Correct Answer:

(c) 90. Total 120; those ≤ 23,000 begin with 1 (24) or 21 (6); 120 − 30 = 90.

Q13. A question paper has 10 MCQs, each with 4 choices. If a student answers every question, how many answer sequences are possible?
(a) 4¹⁰

(b) 10⁴

(d) 4!
Correct Answer:

(a) 4¹⁰. Each of the 10 questions independently has 4 options.

Q14. In how many ways can 8 people be seated at a round table if two particular people must sit together?
(a) 5040

(b) 1440

(d) 10080
Correct Answer:

b) 1440.Treat the pair as one unit → (7 − 1)! = 720, then × 2! for the pair = 1440.

Q15. In how many ways can 10 identical balls be distributed among 4 children so that each child gets at least one?
(a) 36

(b) 84

(d) 210
Correct Answer:

(b) 84. Stars and bars: C(10 − 1, 4 − 1) = C(9,3) = 84.

Q16. A college offers 6 subjects and a student must choose 4. In how many ways?
(a) 10

(b) 12

(d) 20
Correct Answer:

(c) 15. C(6,4) = C(6,2) = 15.

Q17. A password has 3 letters followed by 2 digits with repetition allowed. How many passwords are possible?
(a) 26³ × 10²

(b) 26! × 10!

(d) 26C3 × 10C2
Correct Answer:

(a) 26³ × 10². Each letter slot has 26 options and each digit slot 10, all independent.

Q18. In how many ways can 6 people be seated in a row if two particular people must not sit together?
(a) 240

(b) 480

(d) 720
Correct Answer:

(b) 480. Total 6! = 720; together (treat as block) = 5!×2! = 240; so 720 − 240 = 480.

Q19. In how many ways can 5 red, 4 blue and 3 green balls (identical within each colour) be arranged in a row?
(a) 27720

(b) 13860

(d) 924
Correct Answer

(a) 27720. 12! / (5! × 4! × 3!) = 27720.

Q20. From 8 players, in how many ways can a captain and a vice-captain be chosen?
(a) 28

(b) 64

(d) 16
Correct Answer:

c) 56.Order matters: 8×7 = 56.

Once you are comfortable here, test yourself under time pressure with IPMAT previous year question papers to see how these exact ideas appear in the real exam.

30 More IPMAT Permutation and Combination Practice Questions (with Answers)

These 30 additional questions extend the bank across every sub-topic the IPMAT tests — digit problems, word arrangements, committees, circular seating, and counting with repetition.

Every answer below has been verified. Drill sibling topics alongside these using IPMAT arithmetic questions and how to solve algebra questions for IPMAT.

Q21. In how many ways can all letters of the word "MATHS" be arranged?
(a) 24

(b) 60

(d) 720
Correct Answer: (c) 120. 5! = 120.

Q22. In how many ways can 3 students be chosen from a group of 10?
(a) 30

(b) 120

(d) 1000
Correct Answer: (b) 120. C(10,3) = 120.

Q23. How many 4-digit numbers can be formed using digits 1–9 without repetition?
(a) 3024

(b) 6561

(d) 2016
Correct Answer: (a) 3024. 9×8×7×6 = 3024.

Q24. In how many ways can the letters of the word "DELHI" be arranged?
(a) 60

(b) 120

(d) 720
Correct Answer: (b) 120. 5 distinct letters → 5! = 120.

Q25. A committee of 3 is chosen from 5 men and 4 women (any members). In how many ways?
(a) 84

(b) 72

(d) 60
Correct Answer: (a) 84. C(9,3) = 84.

Q26. In how many ways can 5 different books be arranged on a shelf?
(a) 25

(b) 120

(d) 720
Correct Answer: (b) 120. 5! = 120.

Q27. If 10 people each shake hands once with every other person, how many handshakes occur?
(a) 90

(b) 100

(d) 55
Correct Answer: (c) 45. C(10,2) = 45.

Q28. How many 3-letter codes can be formed from 26 letters if repetition is allowed?
(a) 15600

(b) 17576

(d) 2600
Correct Answer: (b) 17576. 26³ = 17576.

Q29. In how many ways can 5 people be seated around a round table?
(a) 120

(b) 24

(d) 12
Correct Answer: (b) 24. (5 − 1)! = 24.

Q30. From 8 players, in how many ways can a captain and a vice-captain be selected?
(a) 28

(b) 64

(d) 16
Correct Answer: (c) 56. Order matters: 8×7 = 56.

Q31. How many distinct arrangements are there of the letters of the word "BANANA"?
(a) 720

(b) 120

(d) 360
Correct Answer: (c) 60. 6! / (3! × 2!) = 60.

Q32. How many 2-digit even numbers can be formed from 1, 2, 3, 4, 5 without repetition?
(a) 8

(b) 10

(d) 6
Correct Answer: (a) 8. Unit digit 2 or 4 → 2×4 = 8.

Q33. In how many ways can 11 players be selected from a squad of 15?
(a) 1365

(b) 455

(d) 1287

Correct Answer: (a) 1365. C(15,11) = C(15,4) = 1365.

Q34. How many diagonals does a regular octagon have?
(a) 16

(b) 20

(d) 8
Correct Answer: (b) 20. C(8,2) − 8 = 28 − 8 = 20.

Q35. In how many ways can 4 boys and 4 girls be arranged in a row with genders alternating?
(a) 576

(b) 1152

(d) 40320

Correct Answer: (b) 1152. Two starting patterns × 4! × 4! = 2×24×24 = 1152.

Q36. In how many ways can all letters of the word "EQUATION" be arranged?
(a) 5040

(b) 40320

(d) 362880

Correct Answer: (b) 40320. 8 distinct letters → 8! = 40320.

Q37. How many non-empty subsets does a set of 4 elements have?
(a) 16

(b) 15

(d) 24

Correct Answer: (b) 15. 2⁴ − 1 = 15.

Q38. In how many ways can 3 distinct rings be placed on 4 fingers (a finger may hold more than one; ignore order on a finger)?
(a) 12

(b) 24

(d) 81

Correct Answer: (c) 64. Each of 3 rings has 4 finger choices → 4³ = 64.

Q39. How many 5-digit even numbers can be formed from digits 0–9 without repetition (first digit non-zero)?
(a) 13776

(b) 15120

(d) 12000
Correct Answer: (a) 13776.

Q40. There are 3 routes from A to B and 4 routes from B to C. How many routes are there from A to C via B?
(a) 7

(b) 12

(d) 43
Correct Answer: (b) 12. 3×4 = 12.

Q41. In how many ways can a student choose 2 optional subjects out of 5?
(a) 20

(b) 25

(d) 15

Correct Answer: (c) 10. C(5,2) = 10.

Q42. In how many ways can 3 Maths and 2 Physics books be arranged so books of each subject stay together?
(a) 12

(b) 24

(d) 6

Correct Answer: (b) 24. 2! × 3! × 2! = 24.

Q43. A 6-letter word has 4 distinct consonants and 2 distinct vowels. In how many arrangements are the two vowels always together?
(a) 120

(b) 240

(d) 480

Correct Answer: (b) 240. Treat the vowels as one block → 5! × 2! = 240.

Q44. How many triangles can be formed from 10 points, no three of which are collinear?
(a) 720

(b) 120

(d) 100
Correct Answer: (b) 120. C(10,3) = 120.

Q45. How many 4-digit odd numbers can be formed from 1, 2, 3, 4, 5 without repetition?
(a) 48

(b) 72

(d) 120

Correct Answer: (b) 72. Unit digit odd (3 choices) → 3×4×3×2 = 72.

Q46. In how many ways can 5 distinct prizes be distributed among 3 students (a student may win several)?
(a) 243

(b) 125

(d) 60

Correct Answer: (a) 243. Each of 5 prizes has 3 choices → 3⁵ = 243.

Q47. In how many ways can a person climb 5 stairs taking 1 or 2 steps at a time?
(a) 5

(b) 8

(d) 13

Correct Answer: (b) 8. Follows the Fibonacci pattern (1,2,3,5,8) for n = 5.

Q48. In how many ways can 3 ice-cream scoops be chosen from 8 flavours if repetition is allowed and order is irrelevant?

(a) 56

(b) 120

(d) 336
Correct Answer: (b) 120. Combination with repetition: C(8 + 3 − 1, 3) = C(10,3) = 120.

Q49. How many distinct arrangements are there of the letters of the word "LEVEL"?
(a) 30

(b) 60

(d) 20

Correct Answer: (a) 30. 5! / (2! × 2!) = 30 (L and E each repeat twice).

Q50. In how many ways can 2 cards be drawn from a pack of 52 when order matters?
(a) 1326

(b) 2652

(d) 104

Correct Answer: (b) 2652. 52×51 = 2652.

Importance of Permutation and Combination Questions in IPMAT Exam 2027

Why this topic matters?

Permutation and combination sits at the heart of the counting and probability portion of the syllabus.

Mastering it not only earns direct marks but also makes probability questions far easier, since most probability calculations begin with a counting step.

This is why the topic appears consistently across both IPMAT Indore and IPMAT Rohtak. To see exactly where it fits, review the high weightage IPMAT topics for every section.

Read: How to Improve Your Quantitative Aptitude Skills for IPMAT 2027.

Important Formulas to Solve for Permutation and Combination Questions 2027

To work through the IPMAT Permutation and Combination Practice Questions later in this guide, you first need to be confident with the two core formulas below. For a wider formula refresher, keep the IPMAT maths important formulas guide open beside this page.

Permutation (order matters)

nPr = n! / (n − r)!

Use this when you are arranging r items chosen from n distinct items and the sequence is significant.

Combination (order does not matter)

nCr = n! / [ r! × (n − r)! ] = nPr / r!

Use this when you are selecting r items from n and the arrangement is irrelevant.

Special cases you must memorise

Permutation with repetition allowed: n^r (e.g. PINs, codes)
Arrangement with identical objects: n! / (p! × q! × …), where p, q… are the counts of each repeated item (e.g. letters of a word)
Circular permutation of n objects: (n − 1)!

Check: How to Master Maths for IPMAT 2027.

How to Solve IPMAT Questions on Permutation and Combination 2027?

Follow this five-step method to solve every permutation and combination question efficiently.

Step 1: Understand the problem

Read carefully and decide whether the question asks you to arrange (permutation) or select (combination). Watch the keywords: "arrange", "order", "sequence", "seat" point to permutation, while "select", "choose", "group", "team" point to combination.

Step 2: Identify the given values

Find the total number of objects (n) and how many are to be arranged or selected (r). Note special conditions such as whether repetition is allowed, whether identical objects are present, or whether a position is fixed or forbidden.

Step 3: Choose the right formula

For permutation (order matters) use nPr = n! / (n − r)!. For combination (order does not matter) use nCr = n! / [r!(n − r)!].

For repetition use n^r; for identical objects, divide by the factorials of repeats; for circular arrangements use (n − 1)!.

Step 4: Apply the formula and solve

Substitute the values, and cancel common factorial terms before multiplying to avoid large numbers and calculation errors.

Step 5: Verify and cross-check

Confirm you picked the correct model, that every constraint was applied, and that your answer actually appears among the options.

If it does not, you have almost certainly mixed up permutation and combination.

Prepare with: SuperGrads IPMAT Online Coaching and the best IPMAT Quantitative Aptitude books.

Mentor's Shortcut Tips for Permutation and Combination in IPMAT

Run the keyword test first. "Arrange / order / seat" means permutation; "select / choose / team / group" means combination. Decide this before touching a formula.
Use the complement for "at least" problems. Questions like Q1 are almost always faster as total minus none.
Circular tables lose one degree of freedom. Always (n − 1)!, and double it only when a specific pair must sit together.
Repeated letters divide, they do not multiply. For word problems, divide n! by the factorial of each repeated letter's count.
Sanity-check the size. If your answer is not among the options, you have usually confused permutation with combination — re-read the keyword.

Working through this full set of IPMAT Permutation and Combination Practice Questions regularly is the single best way to build the speed and confidence the exam rewards.

Pair the drill with a structured plan from how to master maths for IPMAT.

Key Takeaways

Permutation counts arrangements (order matters); combination counts selections (order does not).
Core formulas: nPr = n!/(n − r)!, nCr = n!/[r!(n − r)!], with n^r, (n − 1)! and multiset division for special cases.
Follow the five-step method: understand, identify values, choose formula, apply, verify.
"At least" questions are usually fastest with complementary counting.
Consistent, timed practice on this topic converts directly into reliable QA marks.

You May Also Find These Useful

IPMAT Maths Syllabus 2027 — the full topic list this chapter belongs to.
Best IPMAT Quantitative Aptitude Books — where to practise more sets.
IPMAT Previous Year Question Papers — see real questions from past papers.
IPMAT Quantitative Aptitude Syllabus — the complete QA blueprint

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