Permutations & Combinations Practice Test (Chapter-8): 100 MCQs for Competitive Exams in India

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.

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.

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.

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.

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!.

Explanation:

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

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.

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.

Explanation:

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

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.

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.

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.

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.

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.

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

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.