Question

Suppose that we want to place 8 non-attacking rooks on a chessboard. In how many ways can we do this if the 16 most ‘northwest' squares must be empty? How about if only the 4 most ‘northwest' squares must be empty?

Answer 1

Total number of squares are 8*8=64,,now of it 16 squares must be empty in the nortwest corner of the board.So lets say we take from the nothern leftmost corner of the board 4 rows and 4 columns,so total 4*4= 16 squares,and then remove that.So now we are left with total 48 squares.

Now to place non attacking rooks ,we cant put any rook in the same row or column of the another rook.

so to place the first rook we have 48 choices for that.

Now for the second one we cannot put it in the row or colum of the first rook and also for the second rook,16 northwest squares arent available,so we are left with 37 choices.As 11 squares are to be omitted for the first rook.

for the third rook,similarly we have 48-20=28 choices,.

for the fouth rook,we have 48-27= 21 choices

for the fifth rook 48-32=16 choices

for the sixth rook,we have 48-39=9 choices

for the seventh rook we have 48-44= 4 choices

for the eight rook we have 48-47=1 choices

So you can come up with a rook configuration after omitting the 16 squares as, by placing the first rook on some column of the fifth row, then the second rook on some other column of the sixth row, and so on.Or you can put the first rook insome rows of the fifth colum , then the second rook on some other row of the sixth column.

so for case 1 ,that is  by placing the first rook on some column of the fifth row, then the second rook on some other column of the sixth row,we have permutation as the number of permutations of the 8 columns= 8!=40320

for the second case,we have permutation as the number of permutations of the 8 rows= 40320.