Question 1

Groups, each containing 3 boys are to be formed out of 5 boys, A, B , C, D and E such that no group can contain both C and D together. What is the maximum number of such different groups?

Accepted Answer

Maximum number of such different groups = ABC , ABD, ABE, BCE, BDE, CEA, DEA = 7

Question 2

In a question paper, there are four multiple choice type question. Each question has five choices with only one choice for its correct answer. What is the total number of ways in which a candidate will not get all the four answers correct?

Accepted Answer

No of multiple choice type questions = 4 Total number of ways = 5 x 5 x 5 x 5 = 625 Number of correct answer = 1 Number of false answers = 625 - 1 = 624

Question 3

In how many difference ways can six players be arranged in a line such that two of them, Abhinav and Manjesh are never together?

Accepted Answer

As, there are six players, so total ways in which they can be arranged = 6 ! ways Also, two particular players,are never together. ∴ Required ways = 6!/2! = 360

Question 4

A, B , C , and D are four towns, any three of which are non-collinear. Then, the number of ways to construct three roads each joining a pair of towns, so that the roads do not form a triangle, is?

Accepted Answer

To construct 2 roads, three towns can be selected out of 4 in 4 x 3 x 2 = 24 ways. Now, if third road goes from the third town to the first town, a triangle is formed and if it goes to the fourth town, a triangle is not formed, So there are 24 ways to form a triangle and 24 ways of avoiding the formation of triangle.

Question 5

Three dice (each having six faces with each face having one number from 1 or 6 ) are rolled. What is the number of possible outcomes such that atleast one dice shows the numbers 2?

Accepted Answer

When three dice are rolled, the number of possible outcomes = 63 = 216 Number of possible outcomes in which 2 does not appear on any dice = 53 = 125 ∴ Number of possible outcomes in which atleast one dice shows 2 = 216 - 125 = 91

Question 6

How many different numbers can be formed from the digits 3, 4, 5, 6 and 7 when repetitions are allowed?

Accepted Answer

Number of 1 digit numbers = 5 Number of 2 digit numbers = 52 = 25 Number of 3 digit numbers = 53 = 125 Number of 4 digit numbers = 54 = 625 Number of 5 digit numbers = 55 = 3125 ∴ Total number of numbers formed with these digits = 5 + 25 + 125 + 625 + 3125 = 3905

Question 7

Find the number of ways of arranging the host and 8 guests at a circular table, so that the host always sits in a particular seat?

Accepted Answer

Total number of persons = 9 Host can sit in a particular seat in one way . Now, remaining positions are defined relative to the host . Hence, the remaining can sit in 8 places in 8P8 = 8! ways. ∴ The number of required arrangements = 8! x 1 = 8! = 8! ways

Question 8

2 Men and 1 women board a bus in which 5 seats are vacant, one of these 5 seats is reserved for ladies. A women may or may not sit on the seat reserved for ladies . In how many different ways can the five seats be occupied by these passengers?

Accepted Answer

Case I :- If lady sets on reserved seat, then 2 men can occupy seats from 4 vacant seats in 4P2 = 4 x 3 = 12 ways Case II :- If lady does not site on reserved seat, then 1 women can occupy a seat from seat in 4 ways, 1 man can occupy a seat from 3 seats in 3 ways, also 1 man left can occupy a seat from remaining two seats in 2 ways. ∴ Total ways = 4 x 3 x 2 = 24 ways Hence, from Case I and case II , total ways = 12 + 24 = 36 ways

Question 9

139 person have signed for an elimination tournament. All players are to be paired up for the first round but because 139 is an odd number one player gets a bye, which promotes him to the second round without actually playing in the first round. The…

Accepted Answer

Required number of member played will be (139 - 1) = 138

Question 10

In a monthly test, the teacher decides that there will be there questions, one each from ex. 7, 8 and 9 of the text book. There are 12 questions in ex-7 , 18 in ex-8 and 9 in ex-9. In how many ways can the three questions be selected?

Accepted Answer

Number of ways selecting 1 question from ex-7 = 12C1 Number of ways selecting 1 question from ex-8 = 18C1 Number of ways selecting 1 question from ex-9 = 9C1 ∴ Total ways = 12C1 x 18C1 x 9C1 = 12 x 18 x 9 = 1944

Question 11

If there are 12 persons in a party and each of them shakes hands with each other, how many handshakes do happen in the party?

Accepted Answer

first person will shake hands with 11 other persons. Seconds person will shake hands with 10 other persons, Third person will shake hands with 9 other person and so on. So , total handshake = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 = 66 OR 2 person are to be chosen from 12 person to have a hand shake ∴ Possible ways = 12C2 = 12! / 2!(12 - 2)! = 66

Question 12

Find the number of ways of preparing a chain with 5 different coloured beads?

Accepted Answer

Neglecting the direction of beads in the chain, number of ways of preparing a chain with 5 different coloured beads = (1/2) x (5 -1)! = (1/2) x 4! = 24/2 = 12

Question 13

Find the number of different ways of forming a committee consisting of 3 men and 3 women from 6 men and 5 women.?

Accepted Answer

The number of ways of selecting 3 men and 3 women out of 6 men and 5 women = 6C3 x 5C3 = 6!/(3! x 3!) + 5/(3! x 2!) = 20 + 10 = 30

Question 14

Two series of a question booklet for an aptitude test are to be given to twelve students. In how many ways can the students be placed in two rows of six each, so that there should be no identical series side by side and that the students sitting one…

Accepted Answer

As, these are two sets of booklets, so number of booklet in each set is 6 and this can be arrange in 6! ways. Also, the other 6 booklets or 2nd set can also be arranged in other 6 students in 6! ways. ∴ Required number of ways = 6! x 6!

Question 15

A department had 8 male and female employees each. A project team involving 3 male and 3 female members needs to be chosen from the department employees. How many different projects teams can be chosen?

Accepted Answer

Total ways = 8C3 x 8C3 = (8 x 7 x 6)/(3 x 2) x (8 x 7 x 6)/(3 x 2) = 56 x 56 = 3136

Question 16

A man has 9 friends, 4 boys and 5 girls. In how many ways can he invite them, if there have to be exactly 3 girls in the invitees?

Accepted Answer

3 girls can be be selected out of 5 girls in 5C3 ways. Since, number of boys to be invited is not given, hence out of 4 boys, he can invite them in (2)4 ways. ∴ Required number of ways 5C3 x (2)4 = 10 x 16 = 160

Question 17

In how many different ways can four books A, B, C and D be arranged one above another in a vertical order such that the books A and B are never in continuous position?

Accepted Answer

The number of arrangement in which A and B are not together = Total number of arrangement - Number of arrangement in which A and B are together = 4! - 3! x 2! = 24 - 12 = 12

Question 18

Boxes numbered 1, 2, 3, 4 and 5 are kept in a row and they are to be filled with either a red or a blue ball such that no two adjacent boxes can be filled with blue balls Then, how different arrangements are possible, given that all balls of a given…

Accepted Answer

Total number of ways filling the 5 boxes numbered as (1, 2, 3, 4 and 5) with either blue or red balls = 25 = 32 Two adjacent boxes with blue balls can be obtained in 4 ways, i.e., (12), (23), (34) and (45). Three adjacent boxes with blue balls can be obtained in 3 ways i.e., (123), (234) and (345). Four adjacent boxes with blue balls can be obtained in 2 ways i.e., (1234) and (2345) and five boxes with blue balls can be got in 1 way. Hence, the total number of ways of filling the boxes such that adjacent boxes have blue balls = (4 + 3 + 2 + 1) = 10 Hence, the number of ways of filling up the boxes such that no two adjacent boxes have blue balls = 32 - 10 = 22

Question 19

A child has four pockets and three marbles. In how many ways, the child can put the marbles in the pockets?

Accepted Answer

The first marble can be put into the pockets in 4 ways, so can the second and third. Thus, the number of ways in which the child can put the marbles = 4 x 4 x 4 = 64 ways .

Question 20

In how many ways, can the letters of the word 'ASSASSINATION' be arranged, so that all the S are together?

Accepted Answer

When All S are taken together, then ASSASSINATION has 10 letters . \ So, 10 letters in total can be arranged in 10 ! ways . [∵ All 'S' are considered as 1] But, here are 3 'A' and 2'I' and 2 'N' . ∴ Required number of ways = 10 ! / (3 ! x 2 ! x 2 !) = 151200

Permutation and Combination Questions