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Please solve 33question urgent help required

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 82=6482=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 72=4972=49 choices.

Then so on, we have 62=3662=36 for the third one, 52=25 for the fourth one, and so on ……

But, however, we have to remember the rooks are not labelled, thus it doesn't matter specifically about a specific rook's position.

Thus, we have a total of   (8!)/8! = 40320 ways.
therefore answer is (B) 8!
 

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another answer

As you have 88 rows and 88 rooks and no two rooks can be on the same row, each row should have exactly one rook.

As you have 88 columns and 88 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 88 different columns such that each of them is covered and none of them repeats. This is the number of permutations of the 88 columns, which is                  8!=8×7×6×5×4×3×2×1=4032

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