There are two identical red, two identical black and two identical white balls. In how many different ways can the balls be placed in the cells (each cell to contain one ball) shown above such that balls of the same colour do not occupy any two consecutiv

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Q: 149 (IAS/2008)
There are two identical red, two identical black and two identical white balls. In how many different ways can the balls be placed in the cells (each cell to contain one ball) shown above such that balls of the same colour do not occupy any two consecutive cells?

question_subject: 

Maths

question_exam: 

IAS

stats: 

0,2,4,0,4,2,0

keywords: 

{'identical white balls': [0, 0, 1, 0], 'consecutive cells': [0, 0, 1, 0], 'balls': [0, 1, 1, 0], 'cells': [0, 0, 0, 1], 'ball': [1, 3, 13, 12], 'identical red': [0, 0, 1, 0], 'cell': [14, 0, 7, 16], 'same colour': [0, 0, 6, 0], 'many different ways': [0, 0, 8, 0]}

The question is about arranging 6 balls (two identical red, two identical black, and two identical white) in such a way that the balls of the same color do not occupy any two consecutive cells.

Option 1 suggests 15 ways of arrangement, but this does not account for all permutations where no two consecutive cells contain balls of the same color.

Option 2 suggests 18 arrangements. This is the correct answer. The basis of this is there are 3! or 6 ways to arrange colors initially, but we have 2 balls of each color, so we multiply by 2^3 or 8. Results in 6*8=48 arrangements. However 2 consecutive balls of same colors can be arranged in 6 ways for each color and must be subtracted from initial 48 leaves 48-(6*3)=18

Option 3 suggests 24 arrangements, but this is more than the possible arrangements with the given condition.

Option 4 suggests 30 arrangements, but this also overestimates the possible arrangements that comply with the condition.

Therefore, Option 2 provides the correct answer.

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