Mathematics Syllabus for JEE Main

Mathematics Syllabus (JEE Main)

Algebra Calculus Coordinate Geometry Sets, Relations & Functions Statistics & Probability Trigonometry Vector & 3D Geometry

Q. How many subsets can be formed from the set {x, y, z, w, v}?

A. 16
B. 32
C. 64
D. 8

Solution

The number of subsets of a set with n elements is 2^n. Here, n = 5, so 2^5 = 32.

Correct Answer: A — 16

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Q. How many subsets does the set A = {a, b, c, d} have?

A. 4
B. 8
C. 16
D. 2

Solution

The number of subsets of a set with n elements is given by 2^n. Here, n = 4, so the number of subsets is 2^4 = 16.

Correct Answer: B — 8

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Q. How many subsets does the set A = {a, b, c} have?

A. 2
B. 3
C. 4
D. 8

Solution

The number of subsets of a set with n elements is 2^n. Here, n = 3, so the number of subsets is 2^3 = 8.

Correct Answer: D — 8

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Q. How many subsets does the set B = {a, b, c, d} have?

A. 4
B. 8
C. 16
D. 2

Solution

The number of subsets of a set with n elements is 2^n. Here, n=4, so the number of subsets is 2^4 = 16.

Correct Answer: B — 8

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Q. How many subsets does the set {a, b, c} have?

A. 2
B. 3
C. 4
D. 8

Solution

The number of subsets of a set with n elements is 2^n. Here, n = 3, so the number of subsets is 2^3 = 8.

Correct Answer: D — 8

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Q. How many subsets of the set H = {x, y} are there that do not contain the element y?

A. 1
B. 2
C. 3
D. 4

Solution

The only subset that does not contain y is {∅} and {x}. Total = 2.

Correct Answer: A — 1

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Q. How many ways can 2 boys and 2 girls be selected from 5 boys and 4 girls?

A. 60
B. 40
C. 50
D. 30

Solution

The number of ways = 5C2 * 4C2 = 10 * 6 = 60.

Correct Answer: A — 60

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Q. How many ways can 2 boys and 2 girls be selected from 6 boys and 4 girls?

A. 60
B. 30
C. 40
D. 50

Solution

The number of ways is C(6,2) * C(4,2) = 15 * 6 = 90.

Correct Answer: A — 60

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Q. How many ways can 3 different books be chosen from a set of 7 books?

A. 35
B. 21
C. 42
D. 28

Solution

The number of ways to choose 3 books from 7 is 7C3 = 7! / (3! * 4!) = 35.

Correct Answer: A — 35

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Q. How many ways can 3 different fruits be chosen from 8 fruits?

A. 56
B. 84
C. 28
D. 36

Solution

The number of ways is C(8,3) = 56.

Correct Answer: B — 84

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Q. How many ways can 3 different fruits be selected from 5 available fruits?

A. 10
B. 15
C. 20
D. 5

Solution

The number of ways to choose 3 from 5 is given by 5C3 = 10.

Correct Answer: B — 15

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Q. How many ways can 3 different letters be chosen from the word 'COMBINATION'?

A. 120
B. 220
C. 60
D. 80

Solution

The number of ways is C(11, 3) = 165.

Correct Answer: C — 60

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Q. How many ways can 3 letters be chosen from the word 'COMBINATION'?

A. 120
B. 60
C. 30
D. 10

Solution

The number of ways to choose 3 letters from 11 distinct letters is 11C3 = 165.

Correct Answer: B — 60

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Q. How many ways can 3 men and 2 women be arranged in a line if the men must be together?

A. 60
B. 120
C. 30
D. 24

Solution

Treat the 3 men as one unit. So, we have 3 units (MMM, W, W). Arrangements = 4! * 3! = 24 * 6 = 144.

Correct Answer: B — 120

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Q. How many ways can 3 red balls and 2 blue balls be arranged in a row?

A. 10
B. 30
C. 60
D. 20

Solution

The arrangements = 5! / (3! * 2!) = 10.

Correct Answer: A — 10

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Q. How many ways can 3 red, 2 blue, and 1 green balls be arranged in a line?

A. 60
B. 120
C. 30
D. 90

Solution

The arrangements = 6! / (3! * 2! * 1!) = 60.

Correct Answer: A — 60

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Q. How many ways can 3 red, 2 blue, and 1 green balls be arranged in a row?

A. 60
B. 120
C. 30
D. 90

Solution

The total arrangements = 6! / (3! * 2! * 1!) = 60.

Correct Answer: A — 60

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Q. How many ways can 4 different books be chosen from a shelf of 10 books?

A. 210
B. 120
C. 240
D. 300

Solution

The number of ways to choose 4 books from 10 is C(10, 4) = 210.

Correct Answer: A — 210

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Q. How many ways can 4 different colored balls be arranged in a line?

A. 16
B. 24
C. 32
D. 48

Solution

The number of arrangements is 4! = 24.

Correct Answer: B — 24

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Q. How many ways can 4 different colored balls be placed in 3 different boxes?

A. 81
B. 64
C. 27
D. 12

Solution

Each ball can go into any of the 3 boxes, so the total ways = 3^4 = 81.

Correct Answer: A — 81

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Q. How many ways can 4 different fruits be selected from a basket of 10 fruits?

A. 210
B. 120
C. 300
D. 150

Solution

The number of ways to choose 4 fruits from 10 is given by 10C4 = 210.

Correct Answer: A — 210

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Q. How many ways can 4 different letters be chosen from the word 'COMBINATION'?

A. 210
B. 126
C. 70
D. 84

Solution

The number of ways to choose 4 letters from 11 distinct letters is 11C4 = 330.

Correct Answer: A — 210

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Q. How many ways can 4 different letters be selected from the word 'COMBINATION'?

A. 210
B. 120
C. 60
D. 30

Solution

The number of ways to choose 4 letters from 11 distinct letters is 11C4 = 330.

Correct Answer: A — 210

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Q. How many ways can 4 different prizes be awarded to 3 students?

A. 12
B. 24
C. 36
D. 48

Solution

The number of ways to award 4 different prizes to 3 students is 3^4 = 81.

Correct Answer: C — 36

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Q. How many ways can 4 different prizes be distributed among 3 students?

A. 81
B. 64
C. 27
D. 12

Solution

Each prize can go to any of the 3 students, so the total ways = 3^4 = 81.

Correct Answer: A — 81

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Q. How many ways can 4 students be selected from a group of 10?

A. 210
B. 120
C. 150
D. 180

Solution

The number of ways is C(10, 4) = 210.

Correct Answer: A — 210

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Q. How many ways can 5 different books be arranged on a shelf?

A. 60
B. 120
C. 100
D. 80

Solution

The number of arrangements of 5 different books is 5! = 120.

Correct Answer: B — 120

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Q. How many ways can 5 different books be selected from a shelf of 10 books?

A. 252
B. 120
C. 200
D. 300

Solution

The number of ways is C(10, 5) = 252.

Correct Answer: A — 252

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Q. How many ways can 5 different letters be arranged such that two specific letters are never together?

A. 48
B. 60
C. 72
D. 96

Solution

Total arrangements = 5! = 120. Arrangements with the two letters together = 4! * 2! = 48. So, arrangements where they are not together = 120 - 48 = 72.

Correct Answer: C — 72

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Q. How many ways can 5 different letters be arranged such that two specific letters are always together?

A. 48
B. 60
C. 120
D. 24

Solution

Treat the two specific letters as one unit. Then, we have 4 units to arrange: 4! = 24. The two letters can be arranged in 2! = 2 ways. Total = 24 * 2 = 48.

Correct Answer: B — 60

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Mathematics Syllabus (JEE Main)

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