Question

Nine squares are chosen at random on a chessboard. What is the probability that they form a square of size $3\times 3$? 

• A. $\displaystyle \frac{9}{\binom{64}{9}}$
• B. $\displaystyle \frac{36}{\binom{64}{9}}$
• C. $\displaystyle \frac{6}{\binom{64}{9}}$
• D. None of these

Hint:

Number of $k\times k$ squares on an $n\times n$ board is $(n-k+1)^2$.

Answer 1

The Correct Option is B

There are $\binom{64}{9}$ ways to choose any $9$ squares from an $8\times 8$ board. A $3\times 3$ block can start in $6$ positions horizontally and $6$ vertically, so there are $6\times 6=36$ such blocks. To “form a $3\times 3$ square,” the chosen $9$ squares must be exactly one of these blocks. Hence \[ P=\frac{\text{favourable}}{\text{total}}=\frac{36}{\binom{64}{9}}. \]

Answer 2

An alternate way to count the favourable outcomes is to use the general formula for placing a smaller square block inside a larger grid, rather than counting the horizontal and vertical positions separately.

1. Option A: This would require only \(9\) favourable placements of the \(3\times 3\) block, but a \(3\times 3\) block can be slid across an \(8\times 8\) board in far more than \(9\) ways, so this undercounts the favourable cases.
2. Option B: For a \(p\times p\) block placed inside an \(n\times n\) grid, the number of possible positions is \((n-p+1)^2\). Here \(n=8\) and \(p=3\), so the block's top-left corner can sit in \((8-3+1)=6\) row-positions and \(6\) column-positions, giving \(6^2=36\) favourable blocks out of \(\binom{64}{9}\) equally likely ways of picking \(9\) squares. This matches the required probability \(\dfrac{36}{\binom{64}{9}}\).
3. Option C: A count of only \(6\) favourable placements would ignore either the row or the column freedom of the block, so it is too small.
4. Option D: Since option B already gives an exact matching expression, "none of these" cannot be correct.

The formula \((n-p+1)^2=36\) confirms that only option B gives the correct favourable count, so the probability is \(\dfrac{36}{\binom{64}{9}}\).

Hence, the correct answer is option B: \(\dfrac{36}{\binom{64}{9}}\).