In how many ways can you place 8 rooks on a 8x8 chessboard so that no row or column contains more than one rook if

 82=6482=64choices for that.

Now, for the second one, we can't be in the row or column of that first one, so leaving us with 72=4972=49 choices.

Then so on, we have 62=3662=36 for the third one, 2525 for the fourth one, and so on ……

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 (8!)28!=40320(8!)28!=40320 ways.

answer: the answer is one of

step-by-step explanation:   we are given to write a proper fraction where the whole is divided into an even number of equal parts less than ten, and an odd number of those parts exist.

an even number less than 10 may be one of 2, 4, 6 or 8. so, the whole may be one of these four evens.

now, an odd number of these parts must exist, so the numerator must be an odd number, one of 1, 3, 5 or 7. it cannot be 9 because then the fraction will become improper.

thus, our fraction may be one of the following

f(x)=2/3x+4/3 is the answer because is whose power in the senate is equal to the power of the speaker of the house?

step-by-step explanation: