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Permutation and Combination — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics Permutation and Combination MCQs with step-by-step solutions covering Fundamental Principle of Counting, Permutations, Combinations, Miscellaneous Applications & Advanced Word Problems. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Fundamental Principle of Counting · easy · numerical
There are 4 roads from town A to town B and 3 roads from town B to town C. In how many ways can a person travel from A to C via B?
A. 16
B. 24
C. 12 ✓ Correct
D. 7
Solution: By the multiplication principle, the journey A to B can be done in 4 ways and B to C in 3 ways, giving 4 × 3 = 12 ways.
Q2 — Fundamental Principle of Counting · easy · numerical
A boy has 5 different shirts and 4 different trousers. In how many ways can he choose one shirt and one trouser to wear?
A. 16
B. 25
C. 20 ✓ Correct
D. 9
Solution: A shirt can be chosen in 5 ways and a trouser in 4 ways, so the outfit can be chosen in 5 × 4 = 20 ways.
Q3 — Fundamental Principle of Counting · easy · numerical
How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6, 7 if no digit is repeated?
A. 840 ✓ Correct
B. 2401
C. 5040
D. 210
Solution: The four places can be filled in 7, 6, 5 and 4 ways respectively, giving 7 × 6 × 5 × 4 = 840 numbers.
Q4 — Fundamental Principle of Counting · easy · numerical
How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 if repetition of digits is allowed?
A. 108
B. 120
C. 216 ✓ Correct
D. 18
Solution: With repetition allowed, each of the three places can be filled in 6 ways, giving 6 × 6 × 6 = 216 numbers.
Q5 — Fundamental Principle of Counting · easy · numerical
How many even 3-digit numbers can be formed from the digits 1, 2, 3, 4, 5 if no digit is repeated?
A. 60
B. 48
C. 12
D. 24 ✓ Correct
Solution: The units place must be 2 or 4 (2 ways); the remaining two places can then be filled in 4 × 3 = 12 ways, giving 2 × 12 = 24 numbers.
Q6 — Fundamental Principle of Counting · easy · numerical
A signal is made by hoisting 2 flags, one above the other, chosen from 5 flags of different colours. How many different signals are possible?
A. 9
B. 25
C. 10
D. 20 ✓ Correct
Solution: The upper flag can be chosen in 5 ways and the lower flag in 4 ways, so there are 5 × 4 = 20 signals.
Q7 — Fundamental Principle of Counting · easy · numerical
A coin is tossed 4 times and the sequence of heads and tails is recorded. How many different outcomes are possible?
A. 16 ✓ Correct
B. 8
C. 4
D. 24
Solution: Each toss has 2 possible results, so the number of outcomes is 2 × 2 × 2 × 2 = 2⁴ = 16.
Q8 — Fundamental Principle of Counting · easy · numerical
Two dice, one red and one blue, are rolled together. How many different outcomes are possible?
A. 12
B. 30
C. 21
D. 36 ✓ Correct
Solution: Each die can land in 6 ways, so the pair of dice gives 6 × 6 = 36 outcomes.
Q9 — Fundamental Principle of Counting · easy · numerical
How many 3-letter codes can be formed from the 26 letters of the English alphabet if letters may be repeated?
A. 17576 ✓ Correct
B. 15600
C. 676
D. 78
Solution: Each of the 3 positions can be filled by any of the 26 letters, so the count is 26 × 26 × 26 = 17576.
Q10 — Fundamental Principle of Counting · easy · numerical
How many 4-digit ATM PINs are possible if each digit can be any digit from 0 to 9 and digits may repeat?
A. 6561
B. 10000 ✓ Correct
C. 9000
D. 5040
Solution: Each of the 4 positions has 10 choices, so the number of PINs is 10 × 10 × 10 × 10 = 10000.
Q11 — Fundamental Principle of Counting · easy · numerical
In how many ways can 4 different letters be dropped into 5 mailboxes, if each letter can go into any mailbox?
A. 120
B. 625 ✓ Correct
C. 1024
D. 20
Solution: Each of the 4 letters can be placed in any of the 5 mailboxes, so the count is 5 × 5 × 5 × 5 = 5⁴ = 625.
Q12 — Fundamental Principle of Counting · easy · numerical
A test has 5 multiple-choice questions, each with 4 options. In how many different ways can a student answer all 5 questions?
A. 1024 ✓ Correct
B. 625
C. 120
D. 20
Solution: Each question can be answered in 4 ways, so all five can be answered in 4 × 4 × 4 × 4 × 4 = 4⁵ = 1024 ways.
Q13 — Permutations · easy · numerical
Find the value of ⁸P₃.
A. 512
B. 56
C. 120
D. 336 ✓ Correct
Solution: ⁸P₃ = 8 × 7 × 6 = 336.
Q14 — 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.
Q15 — 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.
Q16 — 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.
Q17 — 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.
Q18 — 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.
Q19 — 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.
Q20 — 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.
Q21 — 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.
Q22 — 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.
Q23 — 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.
Q24 — 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.
Q25 — Combinations · easy · numerical
The value of ¹²C₄ is:
A. 495 ✓ Correct
B. 792
C. 11880
D. 220
Solution: ¹²C₄ = 12!/(4!·8!) = (12 × 11 × 10 × 9)/24 = 11880/24 = 495.
Q26 — Combinations · easy · numerical
The value of ¹⁵C₂ is:
A. 91
B. 210
C. 105 ✓ Correct
D. 120
Solution: ¹⁵C₂ = (15 × 14)/2 = 210/2 = 105.
Q27 — Combinations · easy · numerical
The value of ²⁰C₁₈ is:
A. 20
B. 380
C. 180
D. 190 ✓ Correct
Solution: Using ⁿCᵣ = ⁿCₙ₋ᵣ, we get ²⁰C₁₈ = ²⁰C₂ = (20 × 19)/2 = 190.
Q28 — Combinations · easy · numerical
In a round-robin tournament every team plays every other team exactly once. If 66 matches are played in all, the number of teams is:
A. 13
B. 11
C. 12 ✓ Correct
D. 66
Solution: ⁿC₂ = 66 gives n(n − 1) = 132 = 12 × 11, so n = 12 teams.
Q29 — Combinations · easy · numerical
If ⁿC₄ = ⁿC₆, then the value of n is:
A. 8
B. 12
C. 10 ✓ Correct
D. 24
Solution: If ⁿCₐ = ⁿC_b with a ≠ b, then a + b = n. So n = 4 + 6 = 10.
Q30 — Combinations · easy · numerical
The value of ⁸C₃ + ⁸C₂ is:
A. 92
B. 84 ✓ Correct
C. 120
D. 64
Solution: By Pascal’s identity ⁸C₃ + ⁸C₂ = ⁹C₃. Indeed 56 + 28 = 84 = ⁹C₃.