Permutations — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics Permutations MCQs with step-by-step solutions (30 questions). Part of Permutation and Combination. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Permutations · easy · numerical
Find the value of ⁸P₃.
A. 512
B. 56
C. 120
D. 336 ✓ Correct
Solution: ⁸P₃ = 8 × 7 × 6 = 336.
Q2 — Permutations · easy · numerical
Find the value of 7! ÷ 5!.
A. 2
B. 56
C. 42 ✓ Correct
D. 21
Solution: 7! ÷ 5! = 7 × 6 = 42, since the factors up to 5! cancel.
Q3 — Permutations · easy · numerical
Find the value of 6! − 4!.
A. 48
B. 720
C. 696 ✓ Correct
D. 2
Solution: 6! = 720 and 4! = 24, so 6! − 4! = 720 − 24 = 696.
Q4 — Permutations · easy · numerical
In how many ways can all the letters of the word TRAIN be arranged?
A. 24
B. 720
C. 120 ✓ Correct
D. 60
Solution: TRAIN has 5 distinct letters, so the arrangements number 5! = 5 × 4 × 3 × 2 × 1 = 120.
Q5 — Permutations · easy · numerical
If ⁿP₂ = 90, find the value of n.
A. 10 ✓ Correct
B. 11
C. 9
D. 45
Solution: ⁿP₂ = n(n − 1) = 90 = 10 × 9, so n = 10.
Q6 — Permutations · easy · numerical
If n! = 5040, find the value of n.
A. 7 ✓ Correct
B. 8
C. 6
D. 9
Solution: 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040, so n = 7.
Q7 — Permutations · easy · numerical
How many different words (with or without meaning) can be formed using all the letters of the word NUMBER?
A. 5040
B. 720 ✓ Correct
C. 360
D. 120
Solution: NUMBER has 6 distinct letters, so the number of arrangements is 6! = 720.
Q8 — Permutations · easy · numerical
In how many ways can the letters of the word APPLE be arranged?
A. 60 ✓ Correct
B. 30
C. 120
D. 24
Solution: APPLE has 5 letters with P repeated twice, so the arrangements number 5! ÷ 2! = 120 ÷ 2 = 60.
Q9 — Permutations · easy · numerical
How many distinct arrangements can be made using all the letters of the word BALLOON?
A. 2520
B. 630
C. 5040
D. 1260 ✓ Correct
Solution: BALLOON has 7 letters with L twice and O twice, so the arrangements number 7! ÷ (2! × 2!) = 5040 ÷ 4 = 1260.
Q10 — Permutations · easy · numerical
In a race with 10 runners, in how many ways can the first, second and third positions be decided (no ties)?
A. 90
B. 1000
C. 720 ✓ Correct
D. 120
Solution: First place can go to any of 10 runners, second to any of 9, third to any of 8, giving 10 × 9 × 8 = 720 ways.
Q11 — Permutations · easy · numerical
In how many ways can 6 people be seated around a circular table?
A. 120 ✓ Correct
B. 60
C. 24
D. 720
Solution: For a circular arrangement of 6 people, the count is (6 − 1)! = 5! = 120.
Q12 — Permutations · easy · numerical
In how many ways can 4 different books be arranged on a shelf?
A. 24 ✓ Correct
B. 12
C. 120
D. 256
Solution: The 4 books can be arranged in 4! = 4 × 3 × 2 × 1 = 24 ways.
Q13 — Permutations · medium · numerical
If (n + 1)! ÷ (n − 1)! = 72, find the value of n.
A. 7
B. 9
C. 8 ✓ Correct
D. 6
Solution: (n + 1)! ÷ (n − 1)! = (n + 1)n = 72 = 9 × 8, so n = 8.
Q14 — Permutations · medium · numerical
If ⁿP₄ = 20 × ⁿP₂, find the value of n.
A. 5
B. 7 ✓ Correct
C. 8
D. 6
Solution: Dividing both sides by ⁿP₂ gives (n − 2)(n − 3) = 20 = 5 × 4, so n − 2 = 5 and n = 7.
Q15 — Permutations · medium · numerical
If ¹⁰Pᵣ = 720, find the value of r.
A. 2
B. 3 ✓ Correct
C. 6
D. 4
Solution: 10 × 9 × 8 = 720, which is a product of 3 consecutive factors starting at 10, so r = 3.
Q16 — Permutations · medium · numerical
How many distinct arrangements can be made using all the letters of the word ASSASSIN?
A. 1680
B. 40320
C. 840 ✓ Correct
D. 420
Solution: ASSASSIN has 8 letters with S four times and A twice, so the arrangements number 8! ÷ (4! × 2!) = 40320 ÷ 48 = 840.
Q17 — Permutations · medium · numerical
How many distinct words (with or without meaning) can be formed by arranging all the letters of the word MATHEMATICS?
A. 39916800
B. 2494800
C. 9979200
D. 4989600 ✓ Correct
Solution: MATHEMATICS has 11 letters with M, A and T each repeated twice, so the count is 11! ÷ (2! × 2! × 2!) = 39916800 ÷ 8 = 4989600.
Q18 — Permutations · medium · numerical
How many distinct arrangements can be made using all the letters of the word COMMITTEE?
A. 90720
B. 45360 ✓ Correct
C. 22680
D. 362880
Solution: COMMITTEE has 9 letters with M, T and E each repeated twice, so the count is 9! ÷ (2! × 2! × 2!) = 362880 ÷ 8 = 45360.
Q19 — Permutations · medium · numerical
How many distinct arrangements can be made using all the letters of the word INDEPENDENCE?
A. 831600
B. 3326400
C. 479001600
D. 1663200 ✓ Correct
Solution: INDEPENDENCE has 12 letters with E four times, N three times and D twice, so the count is 12! ÷ (4! × 3! × 2!) = 479001600 ÷ 288 = 1663200.
Q20 — Permutations · medium · numerical
In how many arrangements of the letters of the word GARDEN do the two vowels always occur together?
A. 720
B. 120
C. 480
D. 240 ✓ Correct
Solution: Treat the vowels A and E as one block: 5 units arrange in 5! = 120 ways, and the vowels arrange within the block in 2! = 2 ways, giving 120 × 2 = 240.
Q21 — Permutations · medium · numerical
In how many ways can the letters of the word ORANGE be arranged so that all the vowels are always together?
A. 144 ✓ Correct
B. 72
C. 720
D. 288
Solution: The vowels O, A, E form one block, giving 4 units that arrange in 4! = 24 ways; the vowels arrange inside the block in 3! = 6 ways, so 24 × 6 = 144.
Q22 — Permutations · medium · numerical
5 boys and 3 girls are to stand in a row so that no two girls are adjacent. In how many ways can they be arranged?
A. 4320
B. 40320
C. 14400 ✓ Correct
D. 2400
Solution: Arrange the boys in 5! = 120 ways; the girls then occupy 3 of the 6 gaps in 6 × 5 × 4 = 120 ways, giving 120 × 120 = 14400.
Q23 — Permutations · medium · numerical
How many arrangements of the letters of the word MONDAY begin with a vowel?
A. 720
B. 240 ✓ Correct
C. 120
D. 480
Solution: The first letter must be O or A (2 ways); the remaining 5 letters arrange in 5! = 120 ways, giving 2 × 120 = 240.
Q24 — Permutations · medium · numerical
How many arrangements of the letters of the word FATHER begin with F and end with R?
A. 24 ✓ Correct
B. 48
C. 720
D. 120
Solution: With F and R fixed at the ends, the remaining 4 letters arrange in the middle in 4! = 24 ways.
Q25 — Permutations · medium · numerical
In how many ways can 8 different flowers be strung to form a garland?
A. 2520 ✓ Correct
B. 5040
C. 1260
D. 40320
Solution: A garland is circular and can be flipped over, so the count is (8 − 1)! ÷ 2 = 5040 ÷ 2 = 2520.
Q26 — Permutations · medium · numerical
How many distinct 6-digit numbers can be formed by rearranging all the digits of the number 233252?
A. 30
B. 720
C. 120
D. 60 ✓ Correct
Solution: The digits are three 2s, two 3s and one 5, so the count is 6! ÷ (3! × 2!) = 720 ÷ 12 = 60.
Q27 — Permutations · medium · numerical
3 mathematics books, 2 physics books and 4 chemistry books (all different) are placed on a shelf so that books of the same subject stay together. In how many ways can this be done?
A. 1728 ✓ Correct
B. 864
C. 288
D. 362880
Solution: The 3 subject blocks arrange in 3! = 6 ways, and within blocks in 3! × 2! × 4! = 6 × 2 × 24 = 288 ways, giving 6 × 288 = 1728.
Q28 — Permutations · medium · numerical
In how many ways can 7 people be arranged in a row if person A must stand somewhere before person B?
A. 5040
B. 720
C. 2520 ✓ Correct
D. 1260
Solution: Of the 7! = 5040 total arrangements, exactly half have A before B, giving 5040 ÷ 2 = 2520.
Q29 — Permutations · medium · numerical
If 5 × ⁴Pᵣ = 6 × ⁵Pᵣ₋₁, find the value of r.
A. 8
B. 4
C. 3 ✓ Correct
D. 2
Solution: Substituting the formulas gives (6 − r)(5 − r) = 6, so r² − 11r + 24 = 0 with roots 3 and 8; since r cannot exceed 4, r = 3.
Q30 — Permutations · medium · numerical
In how many ways can 4 boys and 3 girls be seated in a row so that all the girls sit together?
A. 720 ✓ Correct
B. 144
C. 240
D. 5040
Solution: Treat the 3 girls as one block: the 5 units arrange in 5! = 120 ways and the girls within the block in 3! = 6 ways, giving 120 × 6 = 720.