A board has 16 squares as shown in the figure :
Out of these 16 squares, two squares are chosen at random. The probability that they have no side in common is :

Solution

Maths sir · 3 steps

Count the pairs that do share a side, which is the easy count, and subtract them from all pairs.

1. All pairs. Any 2 of the 16 squares can be chosen, so the total number of pairs is C(16, 2) = 120.
2. Pairs sharing a side. Two squares share a side only when they sit next to each other in the same row or the same column. Each row of 4 squares has 3 side-by-side pairs, so the 4 rows give 4 × 3 = 12. The 4 columns give another 4 × 3 = 12, for a total of 24 pairs.

   P(no common side) = (120 - 24)/120 = 96/120

3. So. 96/120 = 4/5, which is option (b).

The idea to keep. When "no two touching" is hard to count directly, count the touching pairs and subtract. On an m × n grid there are m(n-1) + n(m-1) side-sharing pairs.

Answer from the NTA final answer key. Solution by PrepWiser.