JEE Mathematics 2025 Question with Solution

Given: A board has $16$ squares arranged in a $4 \times 4$ grid. Two squares are chosen at random.

Find: The probability that the chosen squares have no side in common.

Total number of ways to choose any two squares is

$$\binom{16}{2} = \frac{16 \times 15}{2} = 120$$

Now count the pairs that share a side.

In each row of $4$ squares, the number of adjacent horizontal pairs is $3$. Since there are $4$ rows, the number of horizontal adjacent pairs is

$$4 \times 3 = 12$$

Similarly, in each column of $4$ squares, the number of adjacent vertical pairs is $3$. Since there are $4$ columns, the number of vertical adjacent pairs is

$$4 \times 3 = 12$$

So, total pairs sharing a side are

$$12 + 12 = 24$$

Hence, the number of pairs having no side in common is

$$120 - 24 = 96$$

Therefore, the required probability is

$$\frac{96}{120} = \frac{4}{5}$$

So, the correct option is A.

Given: Two squares are selected from a $4 \times 4$ grid.

Find: The probability that the two selected squares do not share a side.

Instead of counting all valid pairs directly, first count all possible pairs and then subtract the pairs that are adjacent.

All possible pairs:

$$\binom{16}{2} = 120$$

Adjacent pairs occur only in two ways:

1. horizontal adjacency
2. vertical adjacency

For horizontal adjacency, each of the $4$ rows contributes $3$ adjacent pairs:

$$4 \times 3 = 12$$

For vertical adjacency, each of the $4$ columns contributes $3$ adjacent pairs:

$$4 \times 3 = 12$$

Thus, total adjacent pairs are

$$12 + 12 = 24$$

So the favorable pairs are

$$120 - 24 = 96$$

Hence,

$$P = \frac{96}{120} = \frac{4}{5}$$

Therefore, the probability that the two chosen squares have no side in common is $\frac{4}{5}$.