<div class="problem">
    <h4>Problem:</h4>
    $8$ rooks are randomly placed on different squares of a chessboard. A rook is said to attack all of the
    squares
    in its row and its column. Compute the probability that every square is occupied or attacked by at least $1$
    rook. You may leave unevaluated binomial coefficients in your answer.
</div>

<div class="solution">
    <h4>Solution:</h4>
    <p>Some observations to note are that... When all the rows and/or all the columns have a rook, then all the
        squares
        are attacked or occupied. So we can make this problem simpler and start off with counting the amount of
        ways
        to
        count the number of ways we can put rooks are in every row and/or column.</p>
    <p>So let's say we start off on the leftmost column. There are \(8\) places for this rook to go. Then we
        move
        onto the next column. There are still
        \(8\) places for this next rook to go, since we aren't caring about the rows in this case. </p>
    <p>So if we continue this pattern for every row, there are $8^8$ ways to arrange the rooks such that they
        cover
        every column. By similar
        reasoning, there are also \(8^8\) ways to arrange the rooks such that they cover every row. So so far we
        have $2
        \cdot 8^8$. But wait! We over counted! In both cases, we over counted where the rooks might be in both
        the
        rows
        and columns! Specifically, we counted that case twice! </p>
    <p>Since there are two cases, we cover all the rows or cover all the columns, we over count the case of
        having
        all the rows AND columns filled twice. We can use PIE or the
        Principle of Inclusion and Exclusion and subtract once, the cases where the rooks are in both the rows
        and
        columns. So if we start from the leftmost column, there are $8$ places to keep our rook. But now we
        don't
        want
        any of our rooks adjacent to each other, so there are $7$ spots in the next column for the next rook.
    </p>
    <p> Similarly,there are $6$ places in the $3$rd column for the $3$rd rook. Then there are $5$ places in the
        4th
        column for the
        $4$th rook. Continuing this pattern gets us a total of $8!$ ways to arrange the rooks. So we have...
        $$8^8 +
        8^8
        - 8! = 2(8^8) - 8! = 33514112.$$ This will serve as the numerator for our probability since it's the
        number
        of
        ways that every square on the board is attacked or occupied. </p>
    <p>
        Our denominator is simply the total ways that we can arrange the $8$ rooks in $64$ squares with no restrictions whatsoever. This will just be... $${64 \choose 8}.$$
    </p>
    <p>
        Our answer follows as... 
        $$\boxed{\frac{33514112}{64 \choose 8}}$$
    </p>
</div>