Five balls of different colours are to be placed in three different boxes such that any box contains at least one ball. What is the maximum number of different ways in which this can be done?

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Submitted by examrobotsa on

Q: 18 (IAS/2007)
Five balls of different colours are to be placed in three different boxes such that any box contains at least one ball. What is the maximum number of different ways in which this can be done?

question_subject: 

Maths

question_exam: 

IAS

stats: 

0,1,6,4,1,1,1

keywords: 

{'different boxes': [0, 0, 1, 0], 'different colours': [0, 0, 2, 0], 'maximum number': [1, 0, 1, 2], 'balls': [0, 1, 1, 0], 'box': [0, 0, 3, 3], 'ball': [1, 3, 13, 12], 'different ways': [0, 0, 5, 0]}

The question involves the distribution of distinct objects into distinct groups.

Option 1 (90 ways): Incorrect. The formula to calculate this scenario is n^r such that n is the number of boxes and r is the number of balls. By this formula, 3^5 would be 243, which is not equal to 90.

Option 2 (120 ways): Correct. Here is the reason: There are 3 boxes, each box has to have at least one ball, hence for the first three balls, they have 3! = 6 ways to be put into the boxes. The remaining 2 balls have 3^2 = 9 choices. Therefore, there are 6*9 = 54 ways for 3 boxes and 3 balls. However, as we have 5 balls and placement of each set having same balls are indistinguishable, we solve it as distribution identical objects into distinct boxes (boxes being distinct ball arrangements here) = 5!/(3!*2!). Therefore, the total number of ways is 5!*54/(3!*2!) = 120.

Option 3 (150 ways): Incorrect. The result that this number represents does not in any way calculate the distribution of distinct items into distinct groups.