Question

In a 8 x 8 chess board, let P be the probability of selecting two squares such that a white queen and a black knight can be placed there in non attacking position. The value of 36 P is (Note: Queen can attack in the same row, column and diagonally. A knight can attack by moving 2 squares horizontally and one square vertically and vice- verse in a single move)

Answer 1

36

Answer 2

Let the total number of squares on the 8x8 chessboard be $N = 8 \times 8 = 64$.

The question asks for the probability of selecting two squares such that a white queen and a black knight can be placed there in non-attacking positions.

The phrasing "selecting two squares" typically implies an unordered selection. The subsequent phrasing "a white queen and a black knight can be placed there in non attacking position" implies that for the selected pair of squares, there exists an assignment of the queen and knight to these squares such that they are non-attacking.

This leads to the interpretation that total outcomes are $\binom{64}{2}$ pairs of squares, and favorable outcomes are pairs $\{s_1, s_2\}$ such that (Q on $s_1$, K on $s_2$ is non-attacking) OR (K on $s_1$, Q on $s_2$ is non-attacking).

Let $s_1=(r_1, c_1), s_2=(r_2, c_2)$.

Q on $s_1$ attacks K on $s_2$ iff $r_1=r_2$ or $c_1=c_2$ or $r_1-c_1 = r_2-c_2$ or $r_1+c_1 = r_2+c_2$.

K on $s_1$ attacks Q on $s_2$ iff $|r_1-r_2|=1$ and $|c_1-c_2|=2$, or $|r_1-r_2|=2$ and $|c_1-c_2|=1$.

We need to find pairs $\{s_1, s_2\}$ such that Q at $s_1$ attacks K at $s_2$ AND K at $s_1$ attacks Q at $s_2$. We will show this number is 0.

Suppose K at $s_1$ attacks Q at $s_2$. Then $|r_1-r_2|=1, |c_1-c_2|=2$ or $|r_1-r_2|=2, |c_1-c_2|=1$.

Case 1: $|r_1-r_2|=1, |c_1-c_2|=2$.

Is $r_1=r_2$? No, $|r_1-r_2|=1$.

Is $c_1=c_2$? No, $|c_1-c_2|=2$.

Is $r_1-c_1 = r_2-c_2$? $r_1-r_2 = c_1-c_2$. $\pm 1 = \pm 2$. Impossible.

Is $r_1+c_1 = r_2+c_2$? $r_1-r_2 = -(c_1-c_2)$. $\pm 1 = \mp 2$. Impossible.

Case 2: $|r_1-r_2|=2, |c_1-c_2|=1$.

Is $r_1=r_2$? No.

Is $c_1=c_2$? No.

Is $r_1-c_1 = r_2-c_2$? $r_1-r_2 = c_1-c_2$. $\pm 2 = \pm 1$. Impossible.

Is $r_1+c_1 = r_2+c_2$? $r_1-r_2 = -(c_1-c_2)$. $\pm 2 = \mp 1$. Impossible.

So, it is indeed true that if K at $s_1$ attacks Q at $s_2$, then Q at $s_1$ does not attack K at $s_2$.

This means the set of squares attacked by a queen at $(r, c)$ and the set of squares attacked by a knight at $(r, c)$ are disjoint.

This means for any square $s_1$, there is no square $s_2$ that is attacked by both Q and K placed at $s_1$.

So the event (Q on $s_1$ attacks K on $s_2$) AND (K on $s_1$ attacks Q on $s_2$) is impossible for any pair of squares $\{s_1, s_2\}$.

The complementary event (Q on $s_1$ is non-attacking K on $s_2$) OR (K on $s_1$ is non-attacking Q on $s_2$) is always true for any pair $\{s_1, s_2\}$.

The number of ways to select two squares is $\binom{64}{2} = 2016$.

For each of these 2016 pairs, it is possible to place the queen and knight in a non-attacking position.

The probability $P = \frac{2016}{2016} = 1$.

The value of $36 P = 36 \times 1 = 36$.