Each of 8 identical balls is to be placed in the squares shown in the figures given in a horizontal direction such that one horizontal row contains 6 balls and the other horizontal row contains 2 balls. In how many maximum different ways can this be done?

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Submitted by examrobotsa on

Q: 133 (IAS/2006)
Each of 8 identical balls is to be placed in the squares shown in the figures given in a horizontal direction such that one horizontal row contains 6 balls and the other horizontal row contains 2 balls. In how many maximum different ways can this be done?

question_subject: 

Logic/Reasoning

question_exam: 

IAS

stats: 

0,2,3,3,2,0,0

keywords: 

{'many maximum different ways': [0, 0, 3, 0], 'identical balls': [0, 0, 3, 0], 'squares': [0, 0, 1, 0], 'balls': [0, 1, 1, 0], 'other horizontal row': [0, 0, 2, 0], 'horizontal row': [0, 0, 2, 0], 'horizontal direction': [0, 0, 2, 2], 'figures': [0, 0, 1, 1]}

The question involves simple permutation of arranging 8 identical balls in two rows; one with 6 locations and another with 2 locations. The 6-ball row has 8 blocks while the 2-ball row only has 4 blocks.

Option 1: The calculation here is incorrect because 8 balls cannot be placed in the two rows in 38 unique ways.

Option 2: This is the correct answer. You can choose two locations for the two balls from the four available in the second row in 4C2=6 ways. Then, you can choose six locations for the remaining six balls from the eight available in the first row in 8C6=28 ways. Multiplying these together gives 6*28=168 ways. Since balls are identical, each arrangement appears 8!/(6!*2!) = 28 times. The total number of unique arrangements is thus 168/28 = 6 ways.

Option 3: Here the calculation seems incorrect because it does not follow the correct permutation principle for identical items

Option 4: Although it involves the number 2 (balls in the second row), this option is incorrect because it doesn`t correctly calculate the total permutations.

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