H (4.84) will be the correct answer
Answer:
The probability is 
Step-by-step explanation:
We can divide the amount of favourable cases by the total amount of cases.
The total amount of cases is the total amount of ways to put 8 rooks on a chessboard. Since a chessboard has 64 squares, this number is the combinatorial number of 64 with 8,
For a favourable case, you need one rook on each column, and for each column the correspondent rook should be in a diferent row than the rest of the rooks. A favourable case can be represented by a bijective function
with A = {1,2,3,4,5,6,7,8}. f(i) = j represents that the rook located in the column i is located in the row j.
Thus, the total of favourable cases is equal to the total amount of bijective functions between a set of 8 elements. This amount is 8!, because we have 8 possibilities for the first column, 7 for the second one, 6 on the third one, and so on.
We can conclude that the probability for 8 rooks not being able to capture themselves is

First seat: 14 candidates to seat
Second seat: 13 candidates to seat
Third seat : 12 candidates
4th seat: 11 candidates
5th seat: 10 candidates
6th seat: 9 candidates
Number of different variations: 14*13*12*11*10*9 = 2,162,160 different ways,
Observe that is 14P6 = 14! / (14-6)! = 14! / 8! = 14*13*12*11*10*9*8! / 8! =
= 14*13*12*11*10*9
Answer: 2,162,160