Permutations and Combinations Practice

Take timed practice tests on Permutations and Combinations for JEE Main and JEE Advanced with session-wise drills, score review, and explanation-led revision.

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Permutations and Combinations Practice Tests

Six 20-question timed sessions plus one 60-question module test. Each question is original and calibrated from the uploaded material pattern without copying PDF wording.

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JEE Main Session 1

4Questions

60s/QTimer

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Preview all 16 questions in Permutations and Combinations Practice Tests (no login required)

Fundamental principle
1. There are 5 routes from A to B and 3 from B to C. Total ways from A to C via B:
- A. 15 (Correct)
- B. 8
- C. 5
- D. 3
Explanation: By the multiplication principle: 5 × 3 = 15 ways.
Permutation formula
2. ⁿPᵣ equals:
- A. n!/(n−r)! (Correct)
- B. n!/r!
- C. n!/(r!(n−r)!)
- D. r!/(n−r)!
Explanation: ⁿPᵣ = n!/(n−r)! = n(n−1)(n−2)...(n−r+1). It counts ordered selections of r items from n distinct items.
Combination formula
3. ⁿCᵣ equals:
- A. n!/(r!(n−r)!) (Correct)
- B. n!/(n−r)!
- C. r!/(n−r)!
- D. n!/r!
Explanation: ⁿCᵣ = n!/(r!(n−r)!). It counts unordered selections (groups) of r items from n distinct items. ⁿCᵣ = ⁿPᵣ/r!.
Basic counting
4. The number of ways to arrange 4 letters from the word MATH is:
- A. ⁴P₄ = 24 (Correct)
- B. ⁴C₄ = 1
- C. 4² = 16
- D. 4³ = 64
Explanation: Arranging all 4 distinct letters: ⁴P₄ = 4! = 24.
Circular permutation
5. The number of ways 6 people can sit around a circular table is:
- A. 5! = 120 (Correct)
- B. 6! = 720
- C. 4! = 24
- D. 6
Explanation: In circular permutations, one person is fixed and the rest are arranged: (n−1)! = (6−1)! = 5! = 120.
With restrictions
6. From 5 men and 3 women, the number of committees of 4 with at least 1 woman is:
- A. 65 (Correct)
- B. 70
- C. 55
- D. 5
Explanation: Total committees of 4 from 8 = ⁸C₄ = 70. Committees with no women = ⁵C₄ = 5. At least 1 woman = 70 − 5 = 65.
Identical items
7. The number of ways to arrange MISSISSIPPI (11 letters: M=1, I=4, S=4, P=2) is:
- A. 11!/(4!4!2!) (Correct)
- B. 11!
- C. 11!/4!
- D. 11!/(4!2!)
Explanation: Arrangements of n objects with repetitions: n!/(n₁!n₂!...). Here: 11!/(4!×4!×2!) = 34650.
Selection with constraints
8. From 10 books, the number of ways to select at least 1 book is:
- A. 2¹⁰ − 1 (Correct)
- B. 2¹⁰
- C. ¹⁰C₁ = 10
- D. 10!
Explanation: Each book can be selected or not: 2¹⁰ total subsets. Subtract the empty set: 2¹⁰ − 1 = 1023.
Rank of a word
9. In dictionary order, the rank of the word CRIME among all arrangements of its letters is:
- A. 22 (Correct)
- B. 32
- C. 30
- D. 36
Explanation: Letters: C, R, I, M, E → sorted: C, E, I, M, R. Words before C__: 0. Fix C, words before CR__: words starting CE, CI, CM = 3×3! = 18. Fix CR, words before CRI__: CRE, CRM... wait sorted after R among remaining {I,M,E}: E,I,M. Before I: E comes first → 1×2! = 2. Fix CRI, words before CRIM: CRI+E = 1×1! = 1. Words before CRIME with CRIM+E: 0. Total = 0+18+2+1+1 = 22+... Let me recount. Remaining letters in CRIME: after sorting: C,E,I,M,R. Starting with C: fixed. Remaining: E,I,M,R. Words starting CE__ = 3! = 6. CI__ = 3! = 6. CM__ = 3! = 6. Now CR__: remaining E,I,M. CRE__ = 2! = 2. CRI__: remaining E,M. CRIME comes before CRIM? Remaining for CRIM_: E → CRIME. Words before CRIME: 6+6+6+2 = 20 + CRIEMX... wait. CR + sorted(E,I,M) = CRE(2!=2 words), CRI(2!=2 words). Under CRI: remaining E,M → CRIME
JEE Main
10. The number of 4-digit numbers with distinct digits that are divisible by 5 is:
- A. 952 (Correct)
- B. 864
- C. 672
- D. 504
Explanation: Divisible by 5 → last digit is 0 or 5. Case 1 (last=0): first digit: 9 choices (1−9), 2nd: 8, 3rd: 7 → 9×8×7 = 504. Case 2 (last=5): first digit: 8 choices (1−9 excluding 5), 2nd: 8, 3rd: 7 → 8×8×7 = 448. Total = 504 + 448 = 952.
Derangement
11. The number of ways 3 letters can be placed in 3 envelopes such that NO letter is in its correct envelope (derangement D₃) is:
- A. 2 (Correct)
- B. 1
- C. 3
- D. 6
Explanation: Derangement formula: Dₙ = n!(1 − 1/1! + 1/2! − 1/3! + ...). D₃ = 3!(1 − 1 + 1/2 − 1/6) = 6(1/3) = 2.
JEE Advanced
12. The number of ways to distribute 10 identical balls into 3 distinct boxes with no box empty is:
- A. ⁹C₂ = 36 (Correct)
- B. ¹⁰C₂ = 45
- C. ¹²C₂ = 66
- D. 10³ = 1000
Explanation: Stars and bars for no empty box: (n−1)C(r−1) = (10−1)C(3−1) = ⁹C₂ = 36.
JEE Advanced
13. The number of ways to select 3 numbers from {1, 2, ..., 10} such that they are in arithmetic progression is:
- A. 16
- B. 20 (Correct)
- C. 15
- D. 10
Explanation: APs from {1..10}: common difference d=1: (1,2,3),(2,3,4),...,(8,9,10) = 8. d=2: (1,3,5),...,(6,8,10) = 6. d=3: (1,4,7),(2,5,8),(3,6,9),(4,7,10) = 4. d=4: (1,5,9),(2,6,10) = 2. d=5: (1,6,11)—invalid. Total = 8+6+4+2 = 20. But wait — we need middle term ≤ 9, and a+2d ≤ 10. For d=1: 8 APs. d=2: 6. d=3: 4. d=4: 2. Total 20. Hmm closest answer is 16 or 20.
Mixed
14. The value of ⁷C₃ + ⁷C₄ is:
- A. ⁸C₄ (Correct)
- B. ⁷C₃
- C. ¹⁴C₃
- D. ⁸C₃
Explanation: Pascal's identity: ⁿCᵣ + ⁿCᵣ₊₁ = ⁿ⁺¹Cᵣ₊₁. So ⁷C₃ + ⁷C₄ = ⁸C₄ = 70.
Mixed
15. In how many ways can a team of 3 boys and 2 girls be chosen from 5 boys and 4 girls?
- A. 60 (Correct)
- B. 40
- C. 120
- D. 20
Explanation: ⁵C₃ × ⁴C₂ = 10 × 6 = 60.
Mixed
16. The number of integers between 1 and 1000 that are divisible by neither 3 nor 5 is:
- A. 533 (Correct)
- B. 466
- C. 400
- D. 600
Explanation: By inclusion-exclusion: divisible by 3 = 333, by 5 = 200, by 15 = 66. Divisible by 3 or 5 = 333+200−66 = 467. Neither = 1000−467 = 533.

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JEE Main Session 1

4Questions

60s/QTimer

+4 / -1Scoring

Preview all 16 questions in Permutations and Combinations Practice Tests (no login required)

Fundamental principle
1. There are 5 routes from A to B and 3 from B to C. Total ways from A to C via B:
- A. 15 (Correct)
- B. 8
- C. 5
- D. 3
Explanation: By the multiplication principle: 5 × 3 = 15 ways.
Permutation formula
2. ⁿPᵣ equals:
- A. n!/(n−r)! (Correct)
- B. n!/r!
- C. n!/(r!(n−r)!)
- D. r!/(n−r)!
Explanation: ⁿPᵣ = n!/(n−r)! = n(n−1)(n−2)...(n−r+1). It counts ordered selections of r items from n distinct items.
Combination formula
3. ⁿCᵣ equals:
- A. n!/(r!(n−r)!) (Correct)
- B. n!/(n−r)!
- C. r!/(n−r)!
- D. n!/r!
Explanation: ⁿCᵣ = n!/(r!(n−r)!). It counts unordered selections (groups) of r items from n distinct items. ⁿCᵣ = ⁿPᵣ/r!.
Basic counting
4. The number of ways to arrange 4 letters from the word MATH is:
- A. ⁴P₄ = 24 (Correct)
- B. ⁴C₄ = 1
- C. 4² = 16
- D. 4³ = 64
Explanation: Arranging all 4 distinct letters: ⁴P₄ = 4! = 24.
Circular permutation
5. The number of ways 6 people can sit around a circular table is:
- A. 5! = 120 (Correct)
- B. 6! = 720
- C. 4! = 24
- D. 6
Explanation: In circular permutations, one person is fixed and the rest are arranged: (n−1)! = (6−1)! = 5! = 120.
With restrictions
6. From 5 men and 3 women, the number of committees of 4 with at least 1 woman is:
- A. 65 (Correct)
- B. 70
- C. 55
- D. 5
Explanation: Total committees of 4 from 8 = ⁸C₄ = 70. Committees with no women = ⁵C₄ = 5. At least 1 woman = 70 − 5 = 65.
Identical items
7. The number of ways to arrange MISSISSIPPI (11 letters: M=1, I=4, S=4, P=2) is:
- A. 11!/(4!4!2!) (Correct)
- B. 11!
- C. 11!/4!
- D. 11!/(4!2!)
Explanation: Arrangements of n objects with repetitions: n!/(n₁!n₂!...). Here: 11!/(4!×4!×2!) = 34650.
Selection with constraints
8. From 10 books, the number of ways to select at least 1 book is:
- A. 2¹⁰ − 1 (Correct)
- B. 2¹⁰
- C. ¹⁰C₁ = 10
- D. 10!
Explanation: Each book can be selected or not: 2¹⁰ total subsets. Subtract the empty set: 2¹⁰ − 1 = 1023.
Rank of a word
9. In dictionary order, the rank of the word CRIME among all arrangements of its letters is:
- A. 22 (Correct)
- B. 32
- C. 30
- D. 36
Explanation: Letters: C, R, I, M, E → sorted: C, E, I, M, R. Words before C__: 0. Fix C, words before CR__: words starting CE, CI, CM = 3×3! = 18. Fix CR, words before CRI__: CRE, CRM... wait sorted after R among remaining {I,M,E}: E,I,M. Before I: E comes first → 1×2! = 2. Fix CRI, words before CRIM: CRI+E = 1×1! = 1. Words before CRIME with CRIM+E: 0. Total = 0+18+2+1+1 = 22+... Let me recount. Remaining letters in CRIME: after sorting: C,E,I,M,R. Starting with C: fixed. Remaining: E,I,M,R. Words starting CE__ = 3! = 6. CI__ = 3! = 6. CM__ = 3! = 6. Now CR__: remaining E,I,M. CRE__ = 2! = 2. CRI__: remaining E,M. CRIME comes before CRIM? Remaining for CRIM_: E → CRIME. Words before CRIME: 6+6+6+2 = 20 + CRIEMX... wait. CR + sorted(E,I,M) = CRE(2!=2 words), CRI(2!=2 words). Under CRI: remaining E,M → CRIME
JEE Main
10. The number of 4-digit numbers with distinct digits that are divisible by 5 is:
- A. 952 (Correct)
- B. 864
- C. 672
- D. 504
Explanation: Divisible by 5 → last digit is 0 or 5. Case 1 (last=0): first digit: 9 choices (1−9), 2nd: 8, 3rd: 7 → 9×8×7 = 504. Case 2 (last=5): first digit: 8 choices (1−9 excluding 5), 2nd: 8, 3rd: 7 → 8×8×7 = 448. Total = 504 + 448 = 952.
Derangement
11. The number of ways 3 letters can be placed in 3 envelopes such that NO letter is in its correct envelope (derangement D₃) is:
- A. 2 (Correct)
- B. 1
- C. 3
- D. 6
Explanation: Derangement formula: Dₙ = n!(1 − 1/1! + 1/2! − 1/3! + ...). D₃ = 3!(1 − 1 + 1/2 − 1/6) = 6(1/3) = 2.
JEE Advanced
12. The number of ways to distribute 10 identical balls into 3 distinct boxes with no box empty is:
- A. ⁹C₂ = 36 (Correct)
- B. ¹⁰C₂ = 45
- C. ¹²C₂ = 66
- D. 10³ = 1000
Explanation: Stars and bars for no empty box: (n−1)C(r−1) = (10−1)C(3−1) = ⁹C₂ = 36.
JEE Advanced
13. The number of ways to select 3 numbers from {1, 2, ..., 10} such that they are in arithmetic progression is:
- A. 16
- B. 20 (Correct)
- C. 15
- D. 10
Explanation: APs from {1..10}: common difference d=1: (1,2,3),(2,3,4),...,(8,9,10) = 8. d=2: (1,3,5),...,(6,8,10) = 6. d=3: (1,4,7),(2,5,8),(3,6,9),(4,7,10) = 4. d=4: (1,5,9),(2,6,10) = 2. d=5: (1,6,11)—invalid. Total = 8+6+4+2 = 20. But wait — we need middle term ≤ 9, and a+2d ≤ 10. For d=1: 8 APs. d=2: 6. d=3: 4. d=4: 2. Total 20. Hmm closest answer is 16 or 20.
Mixed
14. The value of ⁷C₃ + ⁷C₄ is:
- A. ⁸C₄ (Correct)
- B. ⁷C₃
- C. ¹⁴C₃
- D. ⁸C₃
Explanation: Pascal's identity: ⁿCᵣ + ⁿCᵣ₊₁ = ⁿ⁺¹Cᵣ₊₁. So ⁷C₃ + ⁷C₄ = ⁸C₄ = 70.
Mixed
15. In how many ways can a team of 3 boys and 2 girls be chosen from 5 boys and 4 girls?
- A. 60 (Correct)
- B. 40
- C. 120
- D. 20
Explanation: ⁵C₃ × ⁴C₂ = 10 × 6 = 60.
Mixed
16. The number of integers between 1 and 1000 that are divisible by neither 3 nor 5 is:
- A. 533 (Correct)
- B. 466
- C. 400
- D. 600
Explanation: By inclusion-exclusion: divisible by 3 = 333, by 5 = 200, by 15 = 66. Divisible by 3 or 5 = 333+200−66 = 467. Neither = 1000−467 = 533.