Question

Suppose that we put 4 rooks on a standard 8 × 8 chess board so that none of the rooks can capture the others. This means that no two rooks can appear in the same row or column. Furthermore, suppose that we do not put a rook in the upper left corner. How many ways can we do this?

Answer 1

We have to divide it into 2 cases :

1) No rook is placed on the left most column

2) One rook is placed on the left most column

By specifing column and row we give the exact unique location of the rook

After placing first rook no other rook can be placed in that row or column ,ie, 1 column and row are subtracted from the total choice.

Case 1:

1st position = 7(columns) × 8 (rows)

2nd position = 6(columns) × 7( rows)

3rd position = 5 × 6

4th position = 4 × 5

Number of permutations =

Case 2:

1st position in the left column = 7 (rows)

2nd position = 7( columns)× 7(rows)

3rd position = 6 × 6

4th position = 5 × 5

Number of permutations =

Multiplication by 4 is because left position is different from the other 3.

Therefore total number of choices =

In this case I have considered all the rooks to be distinct.

If identical rooks are considered then answer should be divided by 4.