[Joshua0317]

So I am in an introductory probability course at UW Madison. The question I am having difficulties with is:

You have 8 rooks (castles) on a chess board. If you place them all on the board, what is the probability that none of the rooks will capture the other.

Now, I am thinking that the number of events in the sample space is 64C8 because there are 64 spaces and you need to fill 8.

The number of times the event occurs I am thinking is

64*(64-15){since 15 spaces are disqualified for the second piece}*(64-15-13){since 13 new spaces are disqualified; making sure I don't count the original two spaces twice}*(64-15-13-9)... and till I have 8 terms.

When I compute the answer doesn't match my books answer and I don't have a chance to meet with my TA until after the HW is due. Any suggestions?

[ebaines]

I think your approach for the number of ways you can arrange the 8 rooks is fine : it's equivalent to 8*8*7*7*6*6*... *1*1. But instead of using 64C8 in the denominator you should use 64P8. This is because the total mumber of ways to place 8 rooks on a chess board is 64*63*... *57. Using this you get an answer of 9.109E(-6). Does this agree with the book?

[angel_fly_fly]

Hint: No rooks can take each other when each column and row on the chess board contains exactly one rook