Question 1

Question 1: A college offers 5 courses in the morning and 3 in the evening. The number of ways a student can select exactly one course, either in the morning or in the evening is (A) 5 (B) 3 (

Accepted Answer

(C) 8 Hint: Number of ways to select one course from available 8 courses (i.e., 5 courses in the morning and 3 in the evening) = 5 + 3 = 8

Question 2

Question 2: A college has 7 courses in the morning and 3 in the evening. The possible number of choices with the student if he wants to study one course in the morning and one in the evening i

Accepted Answer

(A) 21 Hint: Number of ways to select one morning and one evening course = 7C1 × 3C1 = 21

Question 3

Question 3: In how many ways can 8 Indians and, 4 American and 4 Englishmen can be seated in a row so that all persons of the same nationality sit together? (A) 3! 8! (B) 3! 4! 8! 4! (C) 4! 4!

Accepted Answer

(B) 3! 4! 8! 4! Hint: 8 Indians take their seats in 8! ways, 4 Americans take their seats in 4! ways, 4 Englishmen take their seats in 4! ways. Three groups of Indians, Americans and Englishmen can be permuted in 3! ways. Required number = 3! × 8! × 4! × 4!

Question 4

Question 4: In how many ways can 10 examination papers be arranged so that the best and the worst papers never come together? (A) 9 × 8! (B) 8 × 8! (C) 9 × 9! (D) 8 × 9!

Accepted Answer

(D) 8 × 9! Hint: Arrange 8 papers in 8! ways and two papers in 9 gaps are arranged in 9P2 ways. Required number = 8! 9P2 = 8! × 9 × 8 = 9! × 8

Question 5

Question 5: In how many ways 4 boys and 3 girls can be seated in a row so that they are alternate. (A) 12 (B) 288 (C) 144 (D) 256

Accepted Answer

(C) 144 Hint: B G B G B G B 4 boys take their seats in 4! ways. 3 girls take their seats in 3! ways. Required number = 4! × 3! = 24 × 6 = 144

Question 6

Question 6: Find the number of triangles which can be formed by joining the angular points of a polygon of 8 sides as vertices. (A) 16 (B) 56 (C) 24 (D) 8

Accepted Answer

(B) 56 Hint: A triangle is obtained by joining three vertices. Number of ways of selecting 3 vertices out of 8 vertices = 8C3 = = 56

Question 7

Question 7: A question paper has two parts, A and B, each containing 10 questions. If a student has to choose 8 from part A and 5 from part B, in how many ways can he choose the questions? (A)

Accepted Answer

(D) 11340 Hint: Number of ways to choose 8 questions from Part A and 5 from Part B = 10C8 × 10C5 = 10C2 × 10C5 = 45 × 252 = 11340

Question 8

Question 8: There are 10 persons among whom two are brothers. The total number of ways in which these persons can be seated around a round table so that exactly one person sits between the bro

Accepted Answer

(B) 2! × 8! Hint: Select a person from 8 people (i.e., the people excluding two brothers). This is done in 8 ways. 2 brothers sit adjacent to the selected person on two sides, they may interchange their seats. Remaining 7 people sit in 7! ways Required number = 8 × 2 × 7! = 2! × 8!

Question 9

Question 9: The number of arrangements of the letters of the word BANANA in which two N’s do not appear adjacently is (A) 80 (B) 60 (C) 40 (D) 100

Accepted Answer

(C) 40 Hint: Arrange B, A, A, A in ways. These four letters create 5 gaps in which 2 N are to be filled, this can be done in 5C2 ways, we do not permute those 2N as they are identical. ∴ Required number = × 5C2 = 40

Question 10

Question 10: The number of ways in which 5 male and 2 female members of a committee can be seated around a round table so that the two females are not seated together is (A) 840 (B) 600 (C) 720

Accepted Answer

(D) 480 Hint: 5 males take their seats in 4! ways, creating 5 gaps. In these 5 gaps, 2 females are to be seated. ∴ The number of ways to do this = 5C2 × 2! Required number = 4! × 5C2 × 2! = 480

Question 11

Question 1: Find the value of r if 56Pr+2 : 54Pr-1 = 30800 : 1.

Accepted Answer

Step-by-step solution provided.

Question 12

Question 2: How many words can be formed by writing letters in the word CROWN in a different order?

Accepted Answer

Five Letters of the word CROWN are to be permuted. ∴ Number of different words = 5! = 120

Question 13

Question 3: Find the number of words that can be formed by using all the letters in the word REMAIN. If these words are written in dictionary order, what will be the 40th word?

Accepted Answer

There are 6 letters A, E, I, M, N, R. Number of words that can be formed by using all these letters = 6! = 720 When a word starts with ‘A’, ‘A’ can be arranged in 1 way and the remaining 5 letters can be arranged among themselves in 5! ways. The number of words starting with A = 5! ∴ Similarly, The

Question 14

Question 4: The Capital English alphabet has 11 symmetric letters that appear the same when looked at in a mirror. These letters are A, H, I, M, O, T, U, V, W, X, and Y. How many symmetric thr

Accepted Answer

There are 11 symmetric letters. ∴ Number of 3 Letter passwords = 11P3 = 11 × 10 × 9 = 990

Question 15

Question 5: How many numbers formed using the digits 3, 2, 0, 4, 3, 2, 3 exceed one million?

Accepted Answer

A number that exceeds one million is to be formed from the digits 3, 2, 0, 4, 3, 2, 3. Then the numbers should be any number of 7 digits which can be formed from these digits. Also, among the given numbers 2 is repeated twice and 3 is repeated thrice. ∴ Required number of numbers = Total number of a

Question 16

Question 6: Ten students are to be selected for a project from a class of 30 students. There are 4 students who want to be together either in the project or not in the project. Find the number

Accepted Answer

Ten students are to be selected for a project from a class of 30 students. Case I: If 4 students join the project, then from remaining 26 students, rest of the 6 students are to be selected. Which can be done in 26C6 = = = 230230 ways.

Question 17

Question 7: A student finds 7 books of his interest but can borrow only three books. He wants to borrow the Chemistry part II book only if Chemistry Part I can also be borrowed. Find the numbe

Accepted Answer

There are 7 books of student’s interest, but he can borrow only three books. He wants to borrow the Chemistry part II book only if Chemistry Part I can also be borrowed. Consider the following table: Required number of selections = 5 + 10 = 15

Question 18

Question 8: 30 objects are to be divided into three groups containing 7, 10, 13 objects. Find the number of distinct ways of doing so.

Accepted Answer

First we can select 7 objects out of 30 for the first group in 30C7 ways. Now there are 23 objects left out of which we can select 10 objects for the second group in 23C10 ways. Remaining 13 objects can be selected for the third group in 5C5 ways. ∴ Required number of ways = 30C7 × 23C10 × 13C13 = =

Question 19

Question 9: A student passes an examination if he secures a minimum in each of the 7 subjects. Find the number of ways a student can fail.

Accepted Answer

Every subject a student may pass or fail. ∴ Total number of outcomes = 27 = 128 This number includes one case when the student passes in all subjects. Required number of ways = 128 – 1 = 127

Question 20

Question 10: Nine friends decide to go for a picnic in two groups. One group decides to go by car and the other group decides to go by train. Find the number of different ways of doing so if th

Accepted Answer

Nine friends decide to go for a picnic in two groups and there must be at least 3 friends in each group. Consider the following table:

Question 21

Question 11: A hall has 12 lamps and every lamp can be switched on independently. Find the number of ways of illuminating the hall.

Accepted Answer

Every lamp is either ON or OFF. There are 12 lamps Number of instances = 212 This number includes one case in when all 12 lamps are OFF. ∴ Required Number of ways = 212 – 1 = 4095

Question 22

Question 12: How many quadratic equations can be formed using numbers from 0, 2, 4, 5 as coefficients if a coefficient can be repeated in an equation?

Accepted Answer

A quadratic equation is to be formed using numbers 0, 2, 4, 5 as coefficients and a coefficient can be repeated. Let the quadratic equation be ax2 + bx + c = 0, a ≠ 0 Consider the following table: Number of quadratic equations can be formed = 3 × 4 × 4 = 48

Question 23

Question 13: How many six-digit telephone numbers can be formed if the first two digits are 45 and no digit can appear more than once?

Accepted Answer

There are total of 10 digits. Let the telephone number be 45abcd. There are 8 digits left for the choice of a, b, c, d as repetition is not allowed. Consider the following table: ∴ Required number of numbers formed = 8 × 7 × 6 × 5 = 1680

Question 24

Question 14: A question paper has 6 questions. How many ways does a student have to answer if he wants to solve at least one question?

Accepted Answer

Every question is ‘SOLVED’ or ‘NOT SOLVED’. There are 6 questions. Number of outcomes = 26 This number includes one case when the student solves NONE of the questions. ∴ Required number of ways = 26 – 1 = 64 – 1 = 63

Question 25

Question 15: Find the number of ways of dividing 20 objects into three groups of sizes 8, 7, and 5.

Accepted Answer

First we can select 8 objects our of 20 for the first group in 20C8 ways. Now there are 12 objects left out of which we can select 7 objects for the second group in 12C7 ways. Remaining 5 objects can be selected for the third group in 5C5 ways. ∴ Required number of ways = 20C8 × 12C7 × 5C5 = =

Question 26

Question 16: There are 4 doctors and 8 lawyers in a panel. Find the number of ways for selecting a team of 6 if at least one doctor must be in the team.

Accepted Answer

There are 4 doctors and 8 lawyers in a panel. A team of 6 with at least one doctor is to be formed. We count the number by the INDIRECT method of counting. Number of ways to select a team of 6 people = 12C6 Number of teams with No doctor in any team = 8C6 ∴ Required number of ways = 12C6 – 8C6 = 924

Question 27

Question 17: Four parallel lines intersect another set of five parallel lines. Find the number of distinct parallelograms formed.

Accepted Answer

The first set has 4 parallel lines and another set has 5 parallel lines. To form a parallelogram, we need 2 lines from each set. ∴ Required number of distinct parallelograms formed = 4C2 × 5C2 = 6 × 10 = 60

Question 28

Question 18: There are 12 distinct points A, B, C, …, L, in order, on a circle. Lines are drawn passing through each pair of points. (i) How many lines are there in total? (ii) How many lines p

Accepted Answer

(i) We need two points to draw a line. ∴ Total number of lines = 12C2 = 66

Chapter 3 Permutations and Combination Miscellaneous Exercise 3 Solutions