Latest MBA Aptitude Question SOLUTION: Boxes numbered 1,2,3,4 and 5 are kept in a row and they are filled with either a red or a blue ball, such that two adjacent boxes can be filled with blue balls. Then how many diffe

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(#M40014733) MBA question permutations and combinations, probability Keep an EYE

Boxes numbered 1,2,3,4 and 5 are kept in a row and they are filled with either a red or a blue ball, such that two adjacent boxes can be filled with blue balls. Then how many different arrangements are possible, given that all balls or given colour are exactly identical in all respects?

1) 8
2) 10
3) 15
4) 22

Asked In MBA (11 years ago)

Unsolved Read Solution (4)

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shivam gupta  (7 years ago)

I think the answer should be 13.
consider,
number of ways 0 blue balls(or say 5 balls) can be filled in = 1 way
number of ways 1 blue balls(or say 4 balls) can be filled in = 5C1 ways = 5
similarly, number of ways 2 blue balls(or say 3 balls) not adjacent to each other can be filled in = 4C2 ways=6 ......(this can be acheived as: first put 3 red balls in any 3 boxes and this can be done in 1 way and now since the two balls can be placed in any of the 4 place between red balls)

Similarly, number of ways 3 blue balls be placed so that no two balls are adjacent = 1

So, total ways = 1+5+6+1.

If you have any doubts please try creating a decision tree.

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sweety jain  (9 years ago)

Total number of ways of filling the 5 boxes numbered as (1, 2, 3, 4 and 5) with either blue or red balls = 25 = 32.

Now, let us determine the number of ways of filling the boxes such that the adjacent boxes are filled with blue.

If we decide to have 2 adjacent boxes with blue, it can be done in 4 ways, viz. (12), (23), (34) and (45).

If we decide to have 3 adjacent boxes filled with blue, it can be done in 3 ways, viz. (123), (234) and (345).

If we decide to have 4 adjacent boxes filled with blue, it can be done in 2 ways, viz. (1234) and (2345).

And all 5 boxes can have blue in only 1 way.

Hence, the total number of ways of filling the boxes such that adjacent boxes have blue = (4 + 3 + 2 +1) = 10.

Hence, the number of ways of filling up the boxes such that no two adjacent boxes have blue = 32 - 10 = 22

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Ankit Sinha  (7 years ago)

Answer should be 13 and not 22 or any of the mentioned option.

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shubhanshu  (6 years ago)

Total number of ways of filling the 5 boxes numbered as (1, 2, 3, 4 and 5) with either blue or red balls = 25 = 32.

Now, let us determine the number of ways of filling the boxes such that the adjacent boxes are filled with blue.

If we decide to have 2 adjacent boxes with blue, it can be done in 4 ways, viz. (12), (23), (34) and (45).

If we decide to have 3 adjacent boxes filled with blue, it can be done in 3 ways, viz. (123), (234) and (345).

If we decide to have 4 adjacent boxes filled with blue, it can be done in 2 ways, viz. (1234) and (2345).

And all 5 boxes can have blue in only 1 way.

Hence, the total number of ways of filling the boxes such that adjacent boxes have blue = (4 + 3 + 2 +1) = 10.

Hence, the number of ways of filling up the boxes such that no two adjacent boxes have blue = 32 - 10 = 22

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