Find the number of ways to place 3 rooks on a 5 × 5 chess board so that no two of them attack each other The way I tried this question was that there are $5C_3$ ways to choose a row and a column in which 3 rooks can be placed so I got $5C_3^2$ but it is wrong. The second way I thought was that 1st rook can be placed in any of the first 5 blocks of a row and 2nd rook in 4 ways so total comes out to be $5!$ ways but thats wrong please point out my mistakes

  • $\begingroup$ How did you get 5! ? You're only placing 3 rooks. $\endgroup$
    – cosmo5
    Nov 19 '20 at 22:43

Your partial work of $~\left[\binom{5}{3}\right]^2 = 100$
is the right starting point.

In effect, you have determined that there are 100 possible ways of choosing the pertinent (3 rows) and (3 columns).

Now, you have to consider how many ways are there of placing 3 rooks in a 3 $\times$ 3 chessboard, so that they don't attack each other.

Let $R_i$ denote the rook placed in row $i$.

Then, there are three choices (i.e. columns) that $R_1$ can be placed in, and then two (remaining) columns for $R_2.$

Final answer:

$$\left[\binom{5}{3}\right]^2 \times 3! = 600.$$


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