Permutations & Combinations
Permutation and Combination
star_batch_jee_advanced_2025
Grade None
Question:
The number of ways of selecting two $1 \times 1$ squares from a chess board such that they:
Have a common vertex is 98
Have a common side is 112
Neither have a common vertex nor have a common side is 1806
None of these
Step-by-Step Solution
Key Concept: Systematically count adjacencies (common side), corner-sharing pairs (common vertex), then use complementary counting for pairs with no overlap.
For an $8 \times 8$ chessboard, there are $64$ unit squares total. **Common side:** Two squares share a side if they're adjacent horizontally or vertically. There are $7 \times 8 = 56$ horizontal adjacencies and $8 \times 7 = 56$ vertical adjacencies, giving $112$ pairs. **Common vertex:** Two squares share a vertex if they touch at a corner (including diagonally adjacent). Each internal vertex is shared by 4 squares, each edge vertex by 2, each corner by 1. Counting: $7 \times 7 \times 4 + 7 \times 2 + 7 \times 2 + 1 = 196 + 28 + 1 = 225$ pairs, but subtracting the $112$ with common sides gives $225 - 112 = 113$. Wait, recalculating: $49$ internal vertices give $49 \times 4 = 196$, but divide by $2$ for each pair: $49 \times 2 = 98$ pairs. **Neither:** Total pairs = $\binom{64}{2} = 2016$; neither = $2016 - 112 - 98 = 1806$.
Correct Answer: 1,2,3