Answer:
16,934,400 ways
Step-by-step explanation:
Let n(B) = number of blue rooks
n(B) = 4
Let n(R) = number of red rooks
n(R) = 2
For no two rooks not to attack one another then every rook has to be in a distinct row.
Since there are 2 reds, there can only be 2 spots that the rooks could occupy on the 8x8 board.
So, there are C(8,2) positions the 2 rooks can occupy
Now that we have the number of possible positions that the 2 red rooks can occupy on the entire 8x8 board, we must get the number of possible positions that the 2 red rooks can occupy and cannot attack each other.
The first rook has 8 positions to select.
The second rook has 7 positions
So for the possible outcomes for the red red rooks are:
C(8,2) * 8 * 7
We do the same thing for the blue rooks but this time
There are 8-2 rows left because the 2 rooks has occupied some rows already.
The blue rooks has C(8-2,4) positions to occupy
= C(6,4)
Now that we have the number of possible positions that the 4 blue rooks can occupy on the remainin rows, we must get the number of possible positions that the 4 blue rooks can occupy and cannot attack each other.
The first rook has 6 positions to select.
The second rook has 5 positions
The third has 4 positions
The fourth has 3 positions
So for the possible outcomes for the blue rooks are:
C(6,4) * 6 * 5 * 4 * 3
Total Possible Outcomes = C(8,2) * 8 * 7 * C(6,4) * 6 * 5 * 4 * 3
Total = 16,934,400 ways