The Peaceful Rooks...

08

Let us consider 5 different colored Rooks. White, brown, blue, red, green. On a 6 by 10 board, we can place the white rook anywhere so we have 6 x10 choices. This removes a row and a column. So for the brown rook we have 5x 9 choices. For the blue rook we have 4x8 choices and so on. Altogether 5 distinct Rooks can be placed in 21772800 ways.

Now if the rooks are IDENTICAL, there are 5! ways to permute the positions chosen for the 5 Distinct rooks so there are 21772800 / 5! = 181440 ways to place 5 identical rooks on the board ...

6C5 X 10P5

The rooks wont attack each other if they are in different rows and columns. First we think of the number of permutations possible in the horizontal axis, which is to arrange them in different columns. The horizontal axis can be either 10 or 6. In the first case, the number of permutations will be $10\choose{5}$ $=\frac{10!}{5!\times5!}=252$ . In the second case the number of permutations will be $6\choose{5}$ $=\frac{6!}{5!\times1!}=6$ . After that we only need to multiply this number by the number of permutations possible in the vertical axis, which is to arrange them in different rows, whether it is 6 or 10. For this axis the order matters, so for the first case the number of permutations will be $6!$ , and for the second case the number of permutations will be $10\times9\times8\times7\times6$ . All in all, both cases yield the same answer: $\boxed{181,440}$

Let's choose 5 rows from 10, 10C5. Then, name them as R1, R2, R3, R4, R5. Now, for column, we have to choose 5 column from 6 and lebel them into R1, R2, R3, R4, R5, this is a case of permutation. That can be done in 6P5 ways. So, total answer 10C5*6P5.

Let's picture any sized chessboard. Now, place a rook in a corner of the chessboard, and move it around. Notice that the number of cells the rook can attack are the same for a given chessboard, regardless of where the rook is. The rook can attack exactly one row and one column of the chessboard. Now, to make the solution to this problem simpler, we simply apply the above observation. Since each rook when placed "takes out" both its row and its column, the number of rows and columns available decrease by one every time a rook is placed. Tying up these ideas, we now consider the 6-by-10 chessboard. When one rook is placed, the chessboard is decreased to 5-by-9. When another rook is placed (now, two rooks have been placed), the chessboard is now at a size 4-by-8. For each sized chessboard, we have to count how many cells are available to place the next rook - this is obviously just $rows \times columns$ , and then we apply the Rule of Product. In notation, this is as follows: Number of Arrangements = $(6 \times 10) \times (5 \times 9) \times (4 \times 8) \times (3 \times 7) \times (2 \times 6)$ = $\frac{6!}{1!} \times \frac{10!}{5!} = 10! \times 6$ However, since the rooks are identical, for every one of the arrangements above, the rooks can be rearranged $5!$ times, and so we must divide by this: $\frac{10! \times 6}{5!} = 181440$ possible arrangements.

This is a different method

Imagine a column of $6$ boxes

The number of ways to arrange them is $\frac{6!}{5!}=6$

Imagine a row of 10 boxes

The number of ways to arrange them is $\frac{10!}{2(5!)}$

Since they cannot be on the same row or column, the number of ways should be $\frac{6!}{5!}\times\frac{10!}{2(5!)}$ but no. We forgot that we accounted for the $5!$ twice so the answer really is $2 \times \frac{6!}{5!} \times \frac{10!}{(5!)}=\boxed{181440}$

I times $2$ is because it is the probability of the columns and the rows are interchangeable, and $2!=2$

Refer to this note on Rook Polynomials . Notice that the generalisation for placing $k$ rooks on an $m \times n$ rectangular chessboard is as follows: $r_k(x) = \binom{m}{k} \binom{n}{k} \ k!$

where $r_k(x)$ stands for the $k^{\text{th}}$ rook coefficient of the rook polynomial or the number of ways to place $k$ rooks.