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- Question
- 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
- Correct Answer
- 22
ExplanationTotal 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 - 1. 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
- 2. 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?
Options- A. 320
- B. 160
- C. 80
- D. 200 Discuss
- 3. 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?
Options- A. 112896
- B. 3136
- C. 720
- D. 112 Discuss
- 4. 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
- 5. Find the number of different ways of forming a committee consisting of 3 men and 3 women from 6 men and 5 women.?
Options- A. 30
- B. 20
- C. 10
- D. 25 Discuss
- 6. 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
- 7. In how many ways, can the letters of the word 'ASSASSINATION' be arranged, so that all the S are together?
Options- A. 10!
- B. 14! (4!)
- C. 151200
- D. 3628800 Discuss
- 8. Find the number of ways, in which 12 different beads can be arranged to form a necklace .
Options- A. 11 !/2
- B. 10 !/2
- C. 12 !/2
- D. Couldn't be determined Discuss
- 9. 20 persons were invited to a party. In how many ways, they and the host can be seated at a circular table?
Options- A. 18 !
- B. 19 !
- C. 20 !
- D. Couldn't be determined Discuss
- 10. In how many different ways, 5 boys and 5 girls can sit on a circular table, so that the boys and girls are alternate?
Options- A. 2880
- B. 2800
- C. 2680
- D. 2280 Discuss
Permutation and Combination problems
Search Results
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: 160
Explanation:
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
Correct Answer: 3136
Explanation:
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
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: 30
Explanation:
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
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: 151200
Explanation:
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
Correct Answer: 11 !/2
Explanation:
Number of arrangements of beads = (12 - 1) !, but it is not mentioned that either it is clockwise or anti - clockwise .
So, required number of arrangements = [(12 - 1)! ] / 2 = 11 !/2
Correct Answer: 20 !
Explanation:
Total persons on the circular table
= 20 guests + 1 host = 21
They can be seated in (21 - 1) ! = 20 ! ways .
Correct Answer: 2880
Explanation:
After fixing up one boy on the table, the remaining can be arranged in 4 ! ways, but boys and girls have to be alternate. There will be 5 places, one place each between two boys. These 5 place can be filled by 5 girls in 5 ! ways .
Hence, by the principle of multiplication, the required number of ways = 4 ! x 5 ! = 2880
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