3 rooks are arranged on a 27 times 27 chess board so that no rook is attacking another. How many places can a 4th rook be placed so that it is attacking exactly one other rook on the board
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The fourth rook must be on one of the three rows or three columns occupied by the first three rooks. There are $3\cdot27=81$ squares in the same three rows and another $3\cdot27=81$ squares in the same three columns, but $3\cdot3=9$ of these are in the same three rows and the same three columns, so there are altogether $81+81-9=153$ squares that occupy the same row or column as one of the first three rooks. Three of those squares are occupied by the first three rooks themselves, so there are $153-3=150$ squares from which the fourth rook can attack at least one of the first three.
If it attacks two of the first three rooks, it must attack one along a row and the other along a column. (Why?) This also shows that it can’t attack all three. (Again, why?) From here it should be pretty easy to get to the answer.
Alternatively, you could notice that the answer doesn’t depend on where the first three rooks are placed, so long as they are mutually non-attacking, so you might as well imagine them placed at positions $\langle 1,1\rangle,\langle 2,2\rangle$, and $\langle 3,3\rangle$, where $\langle r,c\rangle$ is the position in row $r$, column $c$. Sketch that picture, and the counting becomes very easy.
Hint: at how many squares is it attacking exactly zero? At how many squares is it attacking exactly two? At how many squares is it attacking exactly three?