To solve the problem, we start by noting the configuration of an 8x8 chessboard, which consists of alternating black and white squares. Each row and each column of this chessboard contains exactly 4 black squares and 4 white squares.
Since we are placing 8 rooks on the chessboard and we want each rook to occupy a black square, we first determine how many black squares are available. On the 8x8 chessboard, there are:
\[ 32 \text{ black squares} \quad (4 \text{ black squares/row} \times 8 \text{ rows}) \]
We seek the probability that all 8 rooks are positioned on black squares. To find this probability, we determine two quantities:
- The total number of ways to place 8 rooks on the chessboard where no two rooks attack each other.
- The total number of ways to place 8 rooks specifically on black squares.
Total Ways to Place 8 Rooks
To place 8 rooks on an 8x8 chessboard such that no two rooks attack each other, we need to select one square from each row and one square from each column. The first rook can be placed in any of the 64 squares, but subsequent rooks have to be placed such that no two occupy the same row or column. The total number of arrangements of 8 rooks is the number of permutations of 8 elements, which is \(8!\).
Ways to Place 8 Rooks on Black Squares
Now, we want to count the ways to place 8 rooks such that each rook is on a black square only. Each row has 4 black squares and we need to pick one square per row such that they are all black. The number of ways to choose one black square from each row (and ensuring that each selected square lies in a different column) is a more restricted combinatorial problem.
The first consideration is to select the specific black squares. There are 4 black squares available in each of the 8 rows, but we need to select 8 positions such that these positions are distributed among the 4 possible columns in each row in such a way that no two rooks are placed in the same column. This is equivalent to choosing 8 positions from 32 black squares, corresponding to valid rook placements.
Each individual arrangement of rooks can be represented by choosing the columns each rook will occupy:
In terms of the arrangement:
- For the first rook in row 1, we have 4 choices of columns to pick a black square.
- For the second row, the rook must also be placed in a black square and in a column not occupied by a previously chosen rook in row 1, giving us 4 choices again and so on for rows 3 through 8.
The placement of rooks can be viewed as assigning columns from the set of 4 black squares while ensuring that we respect the column restrictions:
The number of distinct ways to do this is represented as the 4 choices for the first (which diminishes but remains consistent for columns based on previous establishing). Each of the rows can permute the columns externally, yielding:
After considering all combinatorial choices we get:
\[ 4^8 \]
This gives us the distinct arrangements of rooks.
Final Probability Calculation
Now that we have our totals, the probability \( P \) that all 8 rooks occupy black squares is given by the ratio of the number of favorable outcomes (all rooks on black squares) to the total outcomes (none attacked):
\[ P = \frac{\text{Ways to place rooks on black squares}}{\text{Total ways to place rooks}} = \frac{4^8}{8!} \]
The final expression can be evaluated accordingly, noting that;
\[ P = \frac{65536}{40320} \]
Conclusively, the probability that every rook occupies a black square is
\[ \boxed{\frac{4^8}{8!}} \]
2 answers