A queen on an English chessboard is able to attack in the same row, column and diagonal. The probability that 2 randomly placed queens on an 8 by 8 chessboard will be able to attack each other can be expressed as a/b , where a and b are positive, coprime integers. What is the value of a+b?
NOTE: The 2 queens will not be placed on the same square
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To solve this problem, we need to find the number of ways two queens can be placed on an 8 by 8 chessboard such that they can attack each other.
Let's consider two cases:
Case 1: Two queens are placed in the same row.
In this case, we have 8 possible rows to place the first queen. Once the first queen is placed, there are 7 remaining rows to choose from for the second queen. Therefore, the number of ways to place the queens in the same row is 8 * 7 = 56.
Case 2: Two queens are placed in different rows.
In this case, we have 8 possible rows to place the first queen. Once the first queen is placed, there are 7 remaining rows to choose from for the second queen. However, we need to make sure that the second queen is not in the same column or diagonal as the first queen.
For each row, there are 8 possible columns to place the first queen. Once the first queen is placed, we need to eliminate the columns in the same row and the diagonals from the choices for the second queen.
Since there are 8 possible choices for the first queen's column and 8 possibilities for the second queen's column, the total number of ways to place the queens in different rows is 8 * 7 * 64 = 3584.
To find the total number of ways to place the queens without any restrictions, we sum up the cases:
Total number of ways = 56 + 3584 = 3640.
Finally, we calculate the probability by dividing the number of ways the queens can attack each other by the total number of ways to place the queens:
Probability = (number of ways to attack each other) / (total number of ways) = 56 / 3640 = 7 / 455.
Therefore, a = 7 and b = 455. The sum a + b = 7 + 455 = 462.