Permutation and Combination problems
- 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?
Options- A. 5
- B. 6
- C. 7
- D. 8 Also asked in: Bank Exams
Correct Answer: 7
Explanation:
Maximum number of such different groups = ABC , ABD, ABE, BCE, BDE, CEA, DEA = 7
- 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?
Options- A. 19
- B. 120
- C. 624
- D. 1024 Discuss
Correct Answer: 624
Explanation:
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
- 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?
Options- A. 120
- B. 240
- C. 360
- D. 480 Discuss
Correct Answer: 360
Explanation:
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
- 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?
Options- A. 7
- B. 8
- C. 9
- D. 24 Discuss
Correct Answer: 24
Explanation:
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.
- 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?
Options- A. 36
- B. 81
- C. 91
- D. 116 Discuss
Correct Answer: 91
Explanation:
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
- 6. How many different numbers can be formed from the digits 3, 4, 5, 6 and 7 when repetitions are allowed?
Options- A. 3125
- B. 625
- C. 3905
- D. 125 Discuss
Correct Answer: 3905
Explanation:
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
- 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?
Options- A. 4!
- B. 8!
- C. 6!
- D. 9! Discuss
Correct Answer: 8!
Explanation:
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
- 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?
Options- A. 15
- B. 36
- C. 48
- D. 60 Discuss
Correct Answer: 36
Explanation:
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
- 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 pairing continues on the next round with a bye to any player left over. If the schedule is planned, so that a minimum number of matches is required to determine the champion, the number of matches which must be played is?
Options- A. 136
- B. 137
- C. 138
- D. 139 Discuss
Correct Answer: 138
Explanation:
Required number of member played will be (139 - 1) = 138
- 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?
Options- A. 1944
- B. 2094
- C. 1894
- D. 2194 Discuss
Correct Answer: 1944
Explanation:
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
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