Dimitri places the maximum possible number of Rooks on a 8 × 8 chessboard in such a way that there's no two Rooks that attack each other. In how many ways can he do that?
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Relevant wiki: Permutations without Repetition - Basic
The answer should be 1 0 0 ! (a very huge number and with WolframAlpha I find out it has 158 digits).
But then I tried 8 ! in which 8 is the length of a normal chessboard , and it's correct.
You need to fix the dimensions, Mateo.
Sorry! Copied and pasted the statement from another problem and didn't note that.
Why do you think it is 100! instead of only 8! ?
The board is 8x8, so there are 64 spots that possible for him to place his rook. No matter where he places his second rook it will decrease the size of the board to 7x7, reducing the number of possible spots to 49. Then for his third rook the board will be reduced to 6x6, so he will only have 36 spots. As a result, there are 6 4 ∗ 4 9 ∗ 3 6 ∗ 2 5 ∗ 1 6 ∗ 9 ∗ 4 ∗ 1 = 4 0 3 2 0 spots.
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Relevant wiki: Permutations - Problem Solving
He can place at most 1 rook in each column and row, because if he places 2 or more, they will be attacking eah other. As each row consists of 8 squares, he has 8 possible squares to place the first rook. In the next row, there will be only 7 squares available as one square will be under attack by the first rook. Continuing with this reasoning, for the eighth rook, there is only one square available. Therefore, the answer is simply 8 ! = 4 0 3 2 0 .