Cracking Competitive Exams in India: Permutations and Combinations Practice. In the pursuit of success in competitive exams across India, mastering the art of permutations and combinations is essential. Whether it’s efficiently organizing subject textbooks for exams like SSC and IBPS, forming diverse committees crucial for CRPF and Police Constable exams, or enhancing quantitative aptitude skills for various competitive sectors, these practice questions provide valuable insights for aspiring candidates. These questions are tailored to help you hone your problem-solving abilities and excel in the challenging world of competitive examinations.

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Permutations and Combinations Practice Test-1

20 MCQs 20 Marks 30 Minutes

1 / 20

In how many ways can you choose a president, vice-president, and secretary from a group of 10 people?

Explanation:

To choose a president, vice-president, and secretary, you're looking for permutations. There are 10 choices for the president, 9 choices for the vice-president (since one person has already been chosen as president), and 8 choices for the secretary (as two people have already been chosen). So, the total number of permutations is 10 * 9 * 8 = 720.

3 / 20

A bookstore has 8 different novels and 5 different magazines. In how many ways can you select 3 novels and 2 magazines to read?

Explanation:

This is a combination problem because the order of selection doesn't matter. You need to select 3 novels out of 8, which can be done in C(8, 3) ways, and 2 magazines out of 5, which can be done in C(5, 2) ways. To find the total number of ways to select, multiply these combinations: C(8, 3) * C(5, 2) = 56 * 10 = 1680.

4 / 20

How many ways can you arrange the letters in the word "BOOK"?

Explanation:

To find the number of ways to arrange the letters in the word "BOOK," you use permutations because the order of the letters matters. There are 4 distinct letters in the word, so the number of permutations is 4!.

5 / 20

In a deck of 52 playing cards, how many different 5-card poker hands can be formed that contain a straight (five consecutive cards of any suit)?

Explanation:

To form a poker hand with a straight, you can consider the number of ways to choose the starting card for the straight. There are 10 ways (from A-10 to 10-a). For each starting card, there are 4 suits to choose from, so you have 10 * 4 = 40 possible straights. However, for each straight, there are 5! ways to arrange the cards, so you need to multiply by 5! (the number of ways to arrange 5 cards) to get 40 * 5! = 2,598.

6 / 20

If you have 3 red balls, 2 green balls, and 4 blue balls, how many different ways can you arrange these balls in a row?

Explanation:

To arrange the balls in a row, you use permutations. There are 9 balls in total, so the number of permutations is 9!.

7 / 20

How many ways can you choose a committee of 3 students from a class of 15?

Explanation:

This is a combination problem because the order of selection doesn't matter. You can use the combination formula, which is C(15, 3) = 15! / (3!(15-3)!), to find the number of ways to choose the committee. Calculating it gives you 455, so there are 455 ways to choose a committee of 3 students from a class of 15.

9 / 20

In a bag, there are 8 red marbles and 6 blue marbles. If you randomly draw 4 marbles from the bag, how many different combinations of marbles can you have where exactly 2 are red and 2 are blue?

Explanation:

To find the number of different combinations of marbles where exactly 2 are red and 2 are blue, you can use combinations. Choose 2 red marbles from the 8 available (C(8, 2)), and choose 2 blue marbles from the 6 available (C(6, 2)). Multiply these choices: C(8, 2) * C(6, 2) = 28 * 15 = 420. However, these combinations can be selected in any order, so you divide by 2! (the number of ways to arrange 2 red marbles and 2 blue marbles) to get 420 / 2 = 210.

10 / 20

How many ways can you arrange the letters in the word "BANANA"?

Explanation:

To arrange the letters in "BANANA," you use permutations. However, there are repeated letters (3 'A's and 2 'N's), so you divide by the factorial of the number of repetitions: 6! / (3! * 2!) = 120.

11 / 20

In how many ways can you choose 2 different books from a shelf with 5 different books?

Explanation:

To choose 2 different books from 5 different books, you use combinations. The number of combinations is C(5, 2) = 10.

12 / 20

In a group of 10 people, how many ways can you choose a committee of 3 people and a chairperson from this group if there are 4 men and 6 women, and the committee must include at least one man and one woman?

Explanation:

To choose a committee of 3 people and a chairperson from a group of 10 people with at least one man and one woman, you can calculate it as follows: First, calculate the number of committees with at least one man and one woman. This is the total number of committees (C(10, 3)) minus the committees with only men (C(4, 3)) minus the committees with only women (C(6, 3)). So, C(10, 3) - C(4, 3) - C(6, 3) = 120 - 4 - 20 = 96 committees with at least one man and one woman. Then, for each of these committees, you can choose one of the 3 members as the chairperson in 96 * 3 = 288 ways. Therefore, the total number of ways is 96 * 3 = 288.

13 / 20

How many different license plates can be made using 2 letters followed by 3 digits if repetition is not allowed, and each letter and digit can be used only once?

Explanation:

Since repetition is not allowed, you have 26 choices for the first letter and 25 choices for the second letter. For the digits, you have 10 choices for the first digit, 9 choices for the second digit, and 8 choices for the third digit. To find the total number of license plates, you multiply these choices: 26 * 25 * 10 * 9 * 8 = 5,760.

14 / 20

A boy has 3 library cards and 8 books of his interest in the library. Of these 8, he does not want to borrow chemistry part II unless Chemistry part I is also borrowed) In how many ways can he choose the three books to be borrowed ?

Explanation:

Two possibilities are there : (i) Chemistry part I is available in 8 books with Chemistry part II.
or (ii) Chemistry part II is not available in 8 books but Chemistry part I is available.
Total No. of ways = 1x 1 x 6C1 + 7C3 = 41

16 / 20

How many different 3-letter "words" can you form using the letters A, B, C, D, and E if no letter can be repeated?

Explanation:

To form a 3-letter "word" with no repetition from A, B, C, D, and E, you use permutations. There are 5 choices for the first letter, 4 choices for the second letter (as one letter has already been used), and 3 choices for the third letter. Multiply these choices: 5 * 4 * 3 = 60. However, since the order matters, you divide by 3! (the number of ways to arrange the 3 letters) to get 60 / 6 = 15.

17 / 20

A bookstore has 6 different math textbooks, 4 different physics textbooks, and 5 different chemistry textbooks. In how many ways can you arrange these textbooks on a shelf such that books of the same subject are grouped together?

Explanation:

To arrange the books such that books of the same subject are grouped together, you can treat the books of each subject as a single entity. You now have 3 entities to arrange: (Math, Physics, Chemistry). This can be done in 3! ways. Within each entity, you can arrange the books in different ways. For Math, you have 6! ways, for Physics, you have 4! ways, and for Chemistry, you have 5! ways. Multiply these choices: 3! * 6! * 4! * 5! = 5,040.

18 / 20

In a lottery, there are 50 numbered balls, and you need to select 6 of them. How many different combinations of 6 balls can you choose?

Explanation:

To select 6 balls out of 50, you're looking for combinations. You can calculate this using the combination formula, which is C(50, 6) = 50! / (6!(50-6)!). Calculating it gives you 211,876,220 different combinations of 6 balls you can choose from the lottery.

20 / 20

How many different 2-digit numbers can you form using the digits 1, 2, 3, 4, and 5 if no digit can be repeated?

Explanation:

To form a 2-digit number with no repetition from 1, 2, 3, 4, and 5, you use permutations. There are 5 choices for the first digit and 4 choices for the second digit (as one digit has already been used). Multiply these choices: 5 * 4 = 20.

Topics/Syllabus covered in this Chapter’s Mock Test Series

MCQs questions set for Permutations and Combinations Practice
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