Rooks Attack

Each rook can attack any place on a horizontal or vertical line from the rook. The rooks are thus placed so that none are on the same horizontal or vertical line.

Now, for the fourth rook to attack one of the three other rooks it needs to be on the same horizontal or vertical line, but each line where there is a rook also has two places from which you can attack another rook as well. There is also the spot already occupied by one of the three first rooks. Since the chess board is 27 x 27, each line with a rook on has 27 - 2 - 1 = 24 places to place the fourth rook on.

Since each of the three rooks has one vertical and one horizontal line from which it can be attacked, each rook can be attacked from 24 x 2 = 48 places, and since there were 3 initial rooks there are 24 x 2 x 3 = 144 places that a 4th rook can be placed so that it is attacking exactly one other rook on the board.

When you put a rook on the 27 x 27 chess board, there are 26 x 2 = 52 ways to put another rook so that it will attack the first one. However, for every additional rook in the chess board (third rook, fourth rook, etc), there are 1 x 2 = 2 ways for the second rook to attack it or attack the first rook. Since there are two additional rooks (three in total), the number of ways to put the attacking rook becomes 52 - 2(2) = 48. And because there are three rooks that the attacking rook can attack, 48 x 3 = 144. Therefore, there 144 ways.

Observe that each of the rooks is in a distinct column and a distinct row. This means $3$ rows and $3$ columns are already occupied. Consider placing the fourth rook in the same column as another rook. We have $27-3=24$ choices for the row, because $3$ rows are occupied. We could also place the fourth rook in the same row as that rook in $24$ ways. Because there are $3$ rooks to choose from, our final answer is $(24+24)\cdot3=144$ .

The three rooks that are placed first cannot be placed on the same file or rank. The 3 rooks occupy a file each and a rank each. For the 4th rook to attack one of the already existing rook it needs to be placed on one of the file or rank and to do that, there's 26 places that it can occupy(for a single file or rank). Note that in between this 26 places, 2 will be occupying the same rank, if the attacking rook attacks on the same file, or the same file, if it attacks from a same rank, of another rook. Therefore to attack a only a single rook there's 24 position in the file. The same goes for attacking from a rank. Therefore to attack 1 specified rook, there's 48 position. Since there's 3 rooks that can be attacked, there's 48*3 = 144 position

You can place the three rooks in any squares you want, the number of places to place the 4th rook will be the same. Therefore you can choose the 3 squares in the first row first column, second row second column, and third row third column. Then the 4th rook can go in the first second or third columns anywhere from the fourth row to 27th row, and similarly it can go in the first second or third rows anywhere from the fourth column to 27th column. Thus there are 6*24=144 ways to place the 4th rook.

there are 3 rooks arranged in the chess board so that they are not attacking anyone. no. of ways 4th can be placed so that it will attack only one rook = no. of ways 4th can be placed so that it will attack any no. of rook-no. of ways 4th can be placed so that it will attack 2 rook. (as there is no way that the 4th rook will attack 3 rook because if 4 th rook will attack 3 of them then other three will attack each other.)

no. of ways 4th rook can be placed so that it will attack any no. of rook =

{3*52 -6 \choose 1} =150 (because each will have 26 places in rows and columns and there will be 6 repetition ) no. of ways 4th rook can be placed so that it will attack 2 rook.={6 \choose 1} =6 so required answer will be 150-6=144

There are 6(27-3) ways or 144 ways

Since the first $3$ rooks are not attacking each other, they must each be in a different row and different column. Thus, we must choose a column that contains a rook and a row that does not, or a row that contains a rook and a column that does not. There are $3$ columns containing rooks and $24$ rows not, so there are $3 \times 24 = 72$ ways to choose this. By symmetry, there are also $72$ ways to choose a row with a rook and a column without one. These 2 cases are distinct, so by the rule of sum, there are $144$ ways to choose the positions.