In how many different ways can we place $8$ identical rooks on a chess board so that no two of them attack each other? In how many different ways can we place $8$ identical rooks on a chess board so that no two of them attack each other?
I tried to draw diagrams onto a $8\times8$ square but I'm only getting $16$ ways.  Does that sound right?
Thanks for the help!
 A: For two rooks to be attacking each other, they must either share a row or a column. For two rooks to not
attack each other they must not share a row and they must not share a column.
8 rows and 8 columns need to be chosen once each. A set of eight rows can
be chosen in $\binom{8}{8} = 1$ ways and a set of columns can be chosen in $\binom{8}{8} = 1$ ways.
Ordered pairs can be generated from the two sets in $8!$ ways.
Given the first coordinate of the first ordered pair (can be anything wlog),
there are $\binom{8}{1}$ ways to choose the second coordinate for the first ordered pair.
Given the first coordinate of
the next ordered pair, there are $\binom{7}{1}$ ways to choose the second coordinate.
Because $\binom{n}{1} = n$ the total number of ways to create the coordinates is $n(n−1)(n−2)...(1) = n!$.
In our case, $n = 8$ so there are $8!$ ways to generate valid coordinates or ways place the rooks.
In total, there are $\binom{8}{8} 8! = 8!$ ways to place the rooks.
A: Let's do this piece by piece.
First, let's consider the first rook, we can place it anywhere on the board, thus we have $8^2=64$ choices for that.  
Now, for the second one, we can't be in the row or column of that first one, so leaving us with $7^2=49$ choices.
Then so on, we have $6^2=36$ for the third one, $25$ for the fourth one, and so on $\dots$
But, however, we have to remember the rooks are not labeled, thus it doesn't matter specifically about a specific rook's position.
Thus, we have a total of $\frac{(8!)^2}{8!}=40320$ ways.
A: As you have $8$ rows and $8$ rooks and no two rooks can be on the same row, each row should have exactly one rook.
As you have $8$ columns and $8$ rooks and no two rooks can be on the same column, each column should have exactly one rook.
So you can come up with a rook configuration by placing the first rook on some column of the first row, then the second rook on some other column of the second  row, and so on. The number of configurations is therefore the number of ways you can list the $8$ different columns such that each of them is covered and none of them repeats. This is the number of permutations of the $8$ columns, which is $$8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320.$$
A: We can start off by trying to find a way to apply PIE to this problem. A good question to ask would be, what are the types of formations could the rooks could be in?
Well, the rooks could be occupying every row, every column, or both...AHAAA! We could do (the number of rooks in each row)+(the number of rooks in each column)-(the number of rooks in each column and row).
To find the number combinations for a rook to be in each row, we can simply do $8^8$. This is because there are $8$ possible positions for the rook in the first column, $8$ possible positions for the rook in the second column, and so on. Likewise, there are $8^8$ combinations for the columns.
Now we must find the number of combinations in which there are rooks in every column and row. For the first row, $8$ columns are available, for the second row, $7$ columns are available, and so on until the last row where only $1$ position for the rook is available. This yields $8!$.
Now putting this applying PIE we get $8^8+8^8-8!=2^25-8!=33554432-40320=33514112$
Since we have the numerator figured out, now we just need the denominator. Since there are $64$ squares on a chessboard, and we are choosing $8$ for the rooks, $\binom{64}{8}$ is the denominator.
To conclude we have $\boxed{\frac{33514112}{\binom{64}{8}}}$ as our answer!!!!!
