Dimitri places a Black King and a White Rook on an empty chessboard. If the probability that Dimitri places the King and the Rook on the chessboard such that the King is
NOT
in check (that is, the Rook is not attacking the Black King), can be expressed as
, in which
and
are coprime positive integers, find
.
Details and Assumptions: e.g. If the Rook is on , the , , , , , , , , , , , , and squares are under attack.
The King and the Rook cannot be placed in the same square.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Firstly, note that Dimitri can place the Rook in anyone of the 6 4 squares. As the King and the Rook can't be placed in the same square, Dimitri has 6 3 possible squares to place the Black King. Therefore, there are a total of 6 4 ⋅ 6 3 possible positions.
Secondly, note that, wherever the Rook is placed, will be attacking its whole row and column. This means that the King can't be placed on the same column or row of the Rook. As each row and column is formed by 8 squares, and we are counting twice the square on which the Rook is, there are a total of 6 3 − 1 4 = 4 9 squares on which Dimitri can place the King. Therefore, in each of the 6 4 possible positions of the Rook, the King can be placed on anyone of the 4 9 remaining squares, so the total positions are 4 9 ⋅ 6 4 .
Thus, the probability is 6 3 ⋅ 6 4 4 9 ⋅ 6 4 = 9 7 . Hence, m = 7 and n = 9 , so n − m = 2 .