Permutation and Combination for Competitive Exam Success

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Permutation and Combination

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Permutation and Combination MCQ & Objective Questions

Permutation and Combination is a crucial topic in mathematics that plays a significant role in various exams. Mastering this area can greatly enhance your problem-solving skills and boost your confidence. Practicing MCQs and objective questions on Permutation and Combination helps you familiarize yourself with the types of questions asked in exams, ensuring better preparation and higher scores.

What You Will Practise Here

Fundamental principles of counting
Formulas for permutations and combinations
Applications of permutations in real-life scenarios
Applications of combinations in probability
Common problems involving arrangements and selections
Understanding factorial notation and its significance
Diagrams and visual aids for better comprehension

Exam Relevance

Permutation and Combination is a vital topic in the syllabus of CBSE, State Boards, NEET, and JEE. It frequently appears in various formats, including direct questions, application-based problems, and conceptual MCQs. Students can expect questions that require them to calculate arrangements, selections, or even solve problems involving probability, making it essential to grasp the underlying concepts thoroughly.

Common Mistakes Students Make

Confusing permutations with combinations, especially in word problems.
Neglecting the importance of order in arrangements.
Misapplying formulas due to misunderstanding factorials.
Overlooking restrictions or conditions in selection problems.
Failing to simplify problems before applying formulas.

FAQs

Question: What is the difference between permutations and combinations?
Answer: Permutations consider the order of selection, while combinations do not. For example, arranging books on a shelf is a permutation, whereas selecting books to read is a combination.

Question: How can I improve my speed in solving Permutation and Combination problems?
Answer: Regular practice with MCQs and objective questions will help you identify patterns and improve your calculation speed.

Start solving practice MCQs on Permutation and Combination today to test your understanding and enhance your exam readiness. Remember, consistent practice is the key to mastering this topic and achieving your academic goals!

Q. From a group of 8 people, how many ways can a committee of 3 be formed?

A. 56
B. 24
C. 8
D. 12

Solution

The number of ways to form a committee of 3 from 8 is C(8, 3) = 56.

Correct Answer: A — 56

Q. From a group of 8 people, how many ways can a committee of 4 be formed?

A. 70
B. 56
C. 80
D. 90

Solution

The number of ways to form a committee of 4 from 8 is C(8, 4) = 70.

Correct Answer: A — 70

Q. How many different 4-digit PIN codes can be formed using the digits 0-9 if digits cannot be repeated?

A. 5040
B. 10000
C. 9000
D. 1000

Solution

The number of different 4-digit PIN codes is P(10, 4) = 10! / (10-4)! = 5040.

Correct Answer: A — 5040

Q. How many different 4-digit PIN codes can be formed using the digits 0-9 if repetition is allowed?

A. 10000
B. 1000
C. 5000
D. 9000

Solution

The number of 4-digit PIN codes is 10^4 = 10000.

Correct Answer: A — 10000

Q. How many different 4-digit PIN codes can be formed using the digits 0-9 without repetition?

A. 5040
B. 10000
C. 9000
D. 1000

Solution

The number of different 4-digit PIN codes is P(10, 4) = 10! / (10-4)! = 5040.

Correct Answer: A — 5040

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

A. 60
B. 40
C. 20
D. 30

Solution

The number of ways is C(5, 2) * C(4, 2) = 10 * 6 = 60.

Correct Answer: A — 60

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

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

Solution

The number of ways to choose 3 letters from 'APPLE' (considering 'P' is repeated) is C(5, 3) = 10.

Correct Answer: A — 10

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

A. 56
B. 70
C. 80
D. 90

Solution

The number of ways to choose 3 letters from 8 is C(8, 3) = 56.

Correct Answer: A — 56

Q. How many different ways can 3 students be selected from a group of 10?

A. 120
B. 720
C. 10
D. 100

Solution

The number of ways to select 3 students from 10 is C(10, 3) = 10! / (3! * (10-3)!) = 120.

Correct Answer: A — 120

Q. How many different ways can the letters of the word 'BANANA' be arranged?

A. 60
B. 120
C. 30
D. 20

Solution

The number of arrangements is 6! / (3!) = 20.

Correct Answer: C — 30

Q. How many different ways can the letters of the word 'LEVEL' be arranged?

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

Solution

The number of arrangements of 'LEVEL' is 5! / (2! * 2!) = 30.

Correct Answer: A — 60

Q. How many different ways can you select 2 fruits from 5 different fruits?

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

Solution

The number of ways to choose 2 fruits from 5 is C(5, 2) = 5! / (2!(5-2)!) = 10.

Correct Answer: A — 10

Q. How many ways can 10 different trophies be awarded to 3 different winners?

A. 1000
B. 720
C. 1200
D. 100

Solution

The number of ways to award trophies is P(10, 3) = 10! / (10-3)! = 720.

Correct Answer: A — 1000

Q. How many ways can 2 boys and 3 girls be selected from a group of 5 boys and 6 girls?

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

Solution

The number of ways is C(5, 2) * C(6, 3) = 10 * 20 = 200.

Correct Answer: A — 100

Q. How many ways can 3 different prizes be awarded to 5 students?

A. 60
B. 100
C. 120
D. 30

Solution

The number of ways to award 3 prizes to 5 students is P(5, 3) = 60.

Correct Answer: C — 120

Q. How many ways can 4 different books be arranged on a shelf if 2 specific books must be together?

A. 48
B. 24
C. 36
D. 60

Solution

Treat the 2 specific books as one unit. So, we have 3 units to arrange: (2 books together + 2 other books) = 3! * 2! = 12.

Correct Answer: A — 48

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

A. 24
B. 16
C. 12
D. 8

Solution

The number of ways to arrange 4 different colored balls is 4! = 24.

Correct Answer: A — 24

Q. How many ways can 4 different fruits be selected from 10 available fruits?

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

Solution

The number of ways to select 4 fruits from 10 is C(10, 4) = 10! / (4! × (10-4)!) = 210.

Correct Answer: A — 210

Q. How many ways can 4 people be seated at a round table?

A. 6
B. 12
C. 24
D. 30

Solution

The number of arrangements of 4 people at a round table is (4-1)! = 3! = 6.

Correct Answer: A — 6

Q. How many ways can 4 people be selected from a group of 12?

A. 495
B. 300
C. 400
D. 600

Solution

The number of ways to select 4 people from 12 is C(12, 4) = 495.

Correct Answer: A — 495

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

A. 60
B. 120
C. 100
D. 24

Solution

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

Correct Answer: B — 120

Q. How many ways can 5 different books be chosen from a shelf of 15 books?

A. 3003
B. 5005
C. 1001
D. 1365

Solution

The number of ways to choose 5 books from 15 is C(15, 5) = 3003.

Correct Answer: A — 3003

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

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

Solution

The number of ways to select and arrange 5 books is 5! = 120.

Correct Answer: A — 120

Q. How many ways can 5 different colored balls be placed in 3 different boxes?

A. 243
B. 125
C. 3125
D. 729

Solution

Each ball can go into any of the 3 boxes, so the total ways = 3^5 = 243.

Correct Answer: A — 243

Q. How many ways can 5 different letters be arranged if 2 letters must be together?

A. 48
B. 60
C. 72
D. 80

Solution

Treat the 2 letters as one unit, so we have 4 units to arrange: 4! * 2! = 48.

Correct Answer: C — 72

Q. How many ways can 5 different prizes be awarded to 3 students if each student can receive more than one prize?

A. 243
B. 125
C. 81
D. 64

Solution

Each prize can go to any of the 3 students, so there are 3^5 = 243 ways.

Correct Answer: A — 243

Q. How many ways can 5 different prizes be distributed among 3 students?

A. 243
B. 125
C. 60
D. 30

Solution

Each prize can go to any of the 3 students. Therefore, the total ways = 3^5 = 243.

Correct Answer: A — 243

Q. How many ways can 6 different colored balls be arranged in a row?

A. 720
B. 600
C. 840
D. 480

Solution

The number of arrangements of 6 different colored balls is 6! = 720.

Correct Answer: A — 720

Q. How many ways can 6 people be seated in a round table?

A. 720
B. 120
C. 60
D. 30

Solution

The number of ways to arrange 6 people in a round table is (6-1)! = 5! = 120.

Correct Answer: A — 720

Q. How many ways can 6 people be seated in a row if two specific people must sit next to each other?

A. 120
B. 720
C. 240
D. 60

Solution

Treat the two specific people as one unit. Then, we have 5 units to arrange: 5! = 120. The two people can be arranged in 2! ways. Total = 120 * 2 = 240.

Correct Answer: A — 120

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