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- Question
What is the value of cosec 11?/6?
Options- A. -2/?3
- B. ?2
- C. -2
- D. -?2
- Correct Answer
- -2
- 1. 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
- 2. 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
- 3. 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
- 4. 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
- 5. 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 behind the other should have the same series?
Options- A. 2 x 12C6 x (6!)2
- B. 6! x 6!
- C. 7! x 7!
- D. None of these Discuss
- 6. 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?
Options- A. 9
- B. 12
- C. 14
- D. 18 Discuss
- 7. 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 colour are exactly identical in all respects?
Options- A. 8
- B. 10
- C. 15
- D. 22 Discuss
- 8. A child has four pockets and three marbles. In how many ways, the child can put the marbles in the pockets?
Options- A. 12
- B. 64
- C. 256
- D. 60 Discuss
- 9. In how many ways, a committee of 3 men and 2 women can be formed out of a total of 4 men and 4 women?
Options- A. 15
- B. 16
- C. 20
- D. 24 Discuss
- 10. There is a polygon of 12 sides . How many triangles can be drawn using the vertices of polygon?
Options- A. 200
- B. 220
- C. 240
- D. 260 Discuss
More questions
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
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
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
Correct Answer: 138
Explanation:
Required number of member played will be (139 - 1) = 138
Correct Answer: 6! x 6!
Explanation:
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!
Correct Answer: 12
Explanation:
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
Correct Answer: 22
Explanation:
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
Correct Answer: 64
Explanation:
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 .
Correct Answer: 24
Explanation:
Total number of ways = 4C3 x 4C2 = [4! / {3! x 1!}] x [4! /{2! x 2!}]
= (4 x 4 x 3 x 2 x 1) / (2 x 2)
= 4 x 6
= 24
Correct Answer: 220
Explanation:
Required number of triangles = 12C3 = (12 x 11 x 10) / 6 = 220
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