In how many maximum different ways can 3 identical balls be placed in the 12 squares (each ball to be placed in the exact centre of the squares and only one ball is to be placed in one square) shown in the figure given above such that they do not lie alon

examrobotsa's picture

Submitted by examrobotsa on

Q: 40 (IAS/2007)
In how many maximum different ways can 3 identical balls be placed in the 12 squares (each ball to be placed in the exact centre of the squares and only one ball is to be placed in one square) shown in the figure given above such that they do not lie along the same straight line?

question_subject: 

Logic/Reasoning

question_exam: 

IAS

stats: 

0,2,5,1,3,1,2

keywords: 

{'identical balls': [0, 0, 3, 0], 'many maximum different ways': [0, 0, 3, 0], 'squares': [0, 0, 1, 0], 'square': [0, 0, 0, 1], 'ball': [1, 3, 13, 12], 'same straight line': [0, 0, 1, 0]}

To solve this problem, let`s consider the arrangement of the balls in the squares.

Since each ball needs to be placed in the exact center of a square, we can think of the arrangement as placing the balls in a row of squares. Let`s denote the squares as S1, S2, S3, ..., S12 from left to right.

To ensure that the balls do not lie along the same straight line, we need to consider different possibilities for their placements.

Case 1: All three balls are placed in three different rows.

In this case, we need to select three different rows out of the twelve available rows to place the balls. We can choose three rows out of twelve in 12C3 ways, which is equal to (12!)/(3! * (12-3)!), or 12 * 11 * 10 / (3 * 2 * 1) = 220.

However, we need to subtract the cases where the balls are in a straight line, as per the condition given in the question. In this case, there are four possible straight lines where the balls can lie: S1, S2, S3; S2, S3, S4; S3, S4, S5; and S10, S11, S12. Therefore, we need to subtract these cases from the total.

So, the number of arrangements in Case 1 = 220 - 4 = 216.

Case 2: Two balls are placed in the same row, and the third ball is in a different row.

In this case, we need to select one row out of twelve to place the two balls and another row out of the remaining eleven rows for the third ball. The total number of arrangements in this case is 12 * 11 = 132.

However, we need to subtract the cases where the balls are in a straight line. There are two possible straight lines in this case: S1, S2, S3; and S11, S12, S1. Therefore, we need to subtract these cases from the total.

So, the number of arrangements in Case 2 = 132 - 2 = 130.

Case 3: All three balls are placed in the same row.

In this case, we need to select one row out of twelve to place all three balls. The total number of arrangements in this case is 12.

Now, let`s add up the arrangements from all three cases:

Total arrangements = Case 1 + Case 2 + Case 3

= 216 + 130 + 12

= 358.

However, the question specifically asks for the maximum number of different ways. Looking at the given options, the maximum number mentioned is 216, which matches the number of arrangements in Case 1.

Therefore, the correct answer is Option 4: 216.

Practice this on app