Combinations — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics Combinations 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 — 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.
Q2 — 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.
Q3 — 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.
Q4 — 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.
Q5 — 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.
Q6 — 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.
Q7 — Combinations · easy · numerical
If ⁿCᵣ = 35 and ⁿPᵣ = 210, then the value of r is:
A. 5
B. 3 ✓ Correct
C. 4
D. 2
Solution: ⁿPᵣ = ⁿCᵣ × r!, so r! = 210/35 = 6, giving r = 3.
Q8 — Combinations · easy · numerical
A committee of 4 members is to be formed from 10 people. The number of ways of doing this is:
A. 210 ✓ Correct
B. 5040
C. 252
D. 120
Solution: A committee is an unordered selection: ¹⁰C₄ = (10 × 9 × 8 × 7)/24 = 210.
Q9 — Combinations · easy · numerical
The number of diagonals of a regular hexagon is:
A. 12
B. 15
C. 6
D. 9 ✓ Correct
Solution: Diagonals = ⁶C₂ − 6 = 15 − 6 = 9 (all joins of two vertices minus the 6 sides).
Q10 — Combinations · easy · numerical
8 points are marked on a circle. How many chords can be drawn joining these points?
A. 56
B. 28 ✓ Correct
C. 64
D. 16
Solution: Each pair of points gives one chord: ⁸C₂ = (8 × 7)/2 = 28.
Q11 — Combinations · easy · numerical
At a party, each of the 10 guests shakes hands with every other guest exactly once. The total number of handshakes is:
A. 100
B. 90
C. 45 ✓ Correct
D. 55
Solution: Each handshake is a pair of guests: ¹⁰C₂ = (10 × 9)/2 = 45.
Q12 — Combinations · easy · numerical
In how many ways can 2 books be chosen from a shelf of 7 different books?
A. 42
B. 14
C. 49
D. 21 ✓ Correct
Solution: Order does not matter: ⁷C₂ = (7 × 6)/2 = 21.
Q13 — Combinations · medium · numerical
A committee of 5 is to be formed from 6 men and 4 women so that it contains exactly 2 women. The number of ways is:
A. 126
B. 240
C. 120 ✓ Correct
D. 60
Solution: Choose 2 women and 3 men: ⁴C₂ × ⁶C₃ = 6 × 20 = 120.
Q14 — Combinations · medium · numerical
From 7 men and 5 women, a committee of 5 is to be formed containing at least 3 women. The number of such committees is:
A. 245
B. 210
C. 252
D. 246 ✓ Correct
Solution: Cases: 3W2M = ⁵C₃ × ⁷C₂ = 10 × 21 = 210; 4W1M = 5 × 7 = 35; 5W = 1. Total 210 + 35 + 1 = 246.
Q15 — Combinations · medium · numerical
A box contains 6 different chocolates. In how many ways can a child select at least one chocolate?
A. 64
B. 36
C. 62
D. 63 ✓ Correct
Solution: Each chocolate is taken or not: 2⁶ = 64 selections, minus the empty selection gives 2⁶ − 1 = 63.
Q16 — Combinations · medium · numerical
In how many ways can 4 cards be selected from a standard 52-card pack so that exactly one of them is an ace?
A. 76145
B. 69184 ✓ Correct
C. 17296
D. 270725
Solution: Choose 1 ace from 4 and 3 non-aces from 48: ⁴C₁ × ⁴⁸C₃ = 4 × 17296 = 69184.
Q17 — Combinations · medium · numerical
An urn contains 5 red and 4 blue balls. In how many ways can 3 balls be drawn so that exactly 2 of them are red?
A. 84
B. 40 ✓ Correct
C. 30
D. 60
Solution: Choose 2 red from 5 and 1 blue from 4: ⁵C₂ × ⁴C₁ = 10 × 4 = 40.
Q18 — Combinations · medium · numerical
There are 12 points in a plane, of which 5 are collinear (no other three are collinear). The number of distinct straight lines through these points is:
A. 56
B. 47
C. 57 ✓ Correct
D. 66
Solution: ¹²C₂ − ⁵C₂ + 1 = 66 − 10 + 1 = 57, since the 5 collinear points give one line instead of 10.
Q19 — Combinations · medium · numerical
Out of 10 points in a plane, 4 are collinear and no other three are collinear. The number of triangles formed with these points as vertices is:
A. 116 ✓ Correct
B. 100
C. 112
D. 120
Solution: ¹⁰C₃ − ⁴C₃ = 120 − 4 = 116; the 4 triples of collinear points form no triangle.
Q20 — Combinations · medium · numerical
A polygon has 44 diagonals. The number of its sides is:
A. 22
B. 11 ✓ Correct
C. 10
D. 12
Solution: n(n − 3)/2 = 44 gives n² − 3n − 88 = 0, i.e. (n − 11)(n + 8) = 0, so n = 11.
Q21 — Combinations · medium · numerical
A student must answer 6 out of 10 questions in an exam, but question 1 is compulsory. In how many ways can the questions be chosen?
A. 252
B. 210
C. 84
D. 126 ✓ Correct
Solution: Question 1 is fixed, so choose the remaining 5 from the other 9: ⁹C₅ = 126.
Q22 — Combinations · medium · numerical
A question paper has two parts, each containing 5 questions. A candidate must attempt 6 questions, taking at least 2 from each part. The number of ways of choosing the questions is:
A. 200 ✓ Correct
B. 100
C. 250
D. 210
Solution: Cases (part I, part II): (2,4) = 10 × 5 = 50; (3,3) = 10 × 10 = 100; (4,2) = 5 × 10 = 50. Total 200.
Q23 — Combinations · medium · numerical
If ²ⁿC₃ : ⁿC₃ = 11 : 1, then the value of n is:
A. 8
B. 7
C. 5
D. 6 ✓ Correct
Solution: The ratio simplifies to 4(2n − 1)/(n − 2) = 11, so 8n − 4 = 11n − 22, giving n = 6. Check: ¹²C₃/⁶C₃ = 220/20 = 11.
Q24 — Combinations · medium · numerical
A cricket team of 11 is to be chosen from 15 players, and 2 particular players must always be included. The number of ways of selecting the team is:
A. 286
B. 455
C. 715 ✓ Correct
D. 1365
Solution: With 2 players fixed, choose the remaining 9 from 13: ¹³C₉ = ¹³C₄ = 715.
Q25 — Combinations · medium · numerical
A man has 7 friends. In how many ways can he invite at least 3 of them to a party?
A. 99 ✓ Correct
B. 64
C. 91
D. 128
Solution: ⁷C₃ + ⁷C₄ + ⁷C₅ + ⁷C₆ + ⁷C₇ = 35 + 35 + 21 + 7 + 1 = 99.
Q26 — Combinations · medium · numerical
From 7 consonants and 4 vowels, how many selections of 3 consonants and 2 vowels can be made?
A. 25200
B. 210 ✓ Correct
C. 350
D. 41
Solution: Selections only (no arrangement): ⁷C₃ × ⁴C₂ = 35 × 6 = 210.
Q27 — Combinations · medium · numerical
If ⁿCᵣ₋₁ = 36, ⁿCᵣ = 84 and ⁿCᵣ₊₁ = 126, then the value of n is:
A. 12
B. 8
C. 9 ✓ Correct
D. 10
Solution: The ratios give (n − r + 1)/r = 7/3 and (n − r)/(r + 1) = 3/2; solving, r = 3 and n = 9. Check: ⁹C₂ = 36, ⁹C₃ = 84, ⁹C₄ = 126.
Q28 — Combinations · medium · numerical
A bag has 6 white and 5 black balls. In how many ways can 4 balls be drawn so that at least 2 are black?
A. 210
B. 215 ✓ Correct
C. 330
D. 200
Solution: Cases: 2B2W = ⁵C₂ × ⁶C₂ = 10 × 15 = 150; 3B1W = 10 × 6 = 60; 4B = 5. Total 150 + 60 + 5 = 215.
Q29 — Combinations · medium · numerical
A set of 4 parallel lines intersects another set of 5 parallel lines. The number of parallelograms formed is:
A. 120
B. 20
C. 60 ✓ Correct
D. 40
Solution: A parallelogram needs 2 lines from each family: ⁴C₂ × ⁵C₂ = 6 × 10 = 60.
Q30 — Combinations · medium · numerical
A group of 4 is to be selected from 6 boys and 4 girls so that it contains at least one girl. The number of ways is:
A. 195 ✓ Correct
B. 194
C. 15
D. 210
Solution: Total minus all-boy groups: ¹⁰C₄ − ⁶C₄ = 210 − 15 = 195.