JEE Main 2025 — 23 January, Evening Shift. Previous Year Question.
Problem
A board has 16 squares (arranged as a $4\times4$ grid). Out of these 16 squares, two squares are chosen at random. The probability that they have no side in common is:
Key insight. Rather than trying to directly count all the ways two squares don’t share a side, it’s far easier to count the pairs that do — each row and each column of the grid contributes a small, fixed number of side-adjacent pairs — and subtract that count from the total.
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Approach
Find the total number of ways to choose any 2 squares out of 16. Then count the number of pairs that do share a side, by counting horizontally-adjacent pairs (within each row) and vertically-adjacent pairs (within each column) separately. Subtract this count from the total, and divide by the total to get the probability.
Solution
Step 1 — Find the total number of ways to choose 2 squares
$$\binom{16}{2} = \frac{16\times15}{2} = 120$$
Step 2 — Count horizontally adjacent pairs
Each row has 4 squares in a line, giving 3 adjacent pairs per row (squares 1-2, 2-3, 3-4). With 4 rows:
$$4 \text{ rows} \times 3 \text{ pairs per row} = 12$$
Step 3 — Count vertically adjacent pairs
By the same logic, each column contributes 3 adjacent pairs, with 4 columns:
$$4 \text{ columns} \times 3 \text{ pairs per column} = 12$$
Step 4 — Total pairs sharing a side
$$12+12 = 24$$
Step 5 — Find pairs with no side in common
$$120-24 = 96$$
Step 6 — Compute the probability
$$P(\text{no side in common}) = \frac{96}{120} = \frac{4}{5}$$
Answer
$$\frac{4}{5}$$
Common mistakes
- Forgetting diagonal neighbours don’t count as “sharing a side.” Two squares that touch only at a corner are not excluded by this problem — only squares sharing a full edge count as adjacent.
- Double-counting or missing edge-of-grid effects. It’s tempting to think each of the 16 squares has up to 4 adjacent neighbours, but corner and edge squares have fewer — counting adjacent pairs row-by-row and column-by-column avoids this trap entirely, since each pair is counted exactly once.
Practise next
- A board has 9 squares arranged as a $3\times3$ grid. Two squares are chosen at random. Find the probability that they share no side, using the same row-and-column counting method.
Show answer
$\dfrac23$. Two squares are chosen from nine in $\binom92=36$ ways.
Count the adjacent pairs by direction: each of the $3$ rows contributes $2$ horizontally adjacent pairs, and each of the $3$ columns contributes $2$ vertically adjacent pairs — $6+6=12$ in all.
So pairs sharing no side number $36-12=24$, and the probability is $\dfrac{24}{36}=\dfrac23$. Counting the bad cases by rows and columns is far safer than trying to count the good ones directly.
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