In the figure shown, what is the maximum number of different ways in which 8 identical balls can be placed in the small triangles 1, 2, 3 and 4 such that each triangle contains at least one ball?

examrobotsa's picture
Q: 45 (IAS/2007)
In the figure shown, what is the maximum number of different ways in which 8 identical balls can be placed in the small triangles 1, 2, 3 and 4 such that each triangle contains at least one ball?

question_subject: 

Logic/Reasoning

question_exam: 

IAS

stats: 

0,1,6,1,3,2,1

keywords: 

{'identical balls': [0, 0, 3, 0], 'small triangles': [0, 0, 1, 0], 'triangle': [0, 1, 0, 1], 'ball': [1, 3, 13, 12], 'maximum number': [1, 0, 1, 2], 'figure': [0, 1, 1, 0], 'different ways': [0, 0, 5, 0]}

The question consists of a combinatorial problem where we are asked to distribute 8 identical balls into 4 distinct small triangles, each triangle must contain at least one ball. Here, any given triangle can contain 1, 2, 3, or more up to 8 balls but not 0.

Option 1:"32" is indeed the correct answer. This utilizes the "Stars and Bars" combinatorial principle, where instead of 8 balls, we consider distributing 8-4 = 4 "extra" balls (as one ball is already in each triangle). The number of ways is thus C(4+4-1, 4) = 35. But, we have to deduct cases in which any triangle gets more than 5 (8-3=5) balls, which is C(4,1)*C(3+4-1,3)=3. Thus, final answer = 35-3=32.

Option 2: "35" signifies the total number of ways 8-4 balls can be distributed in 4 boxes, regardless of any further conditions.

Option 3: "44" has no specific reasoning or computations related to this problem.

Option 4: "56" also

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