In how many ways can two red and four blue rooks be placed on an 8-by-8 board so that no two rooks can attack one another?

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

a) 0,0; 1,1; and 2,2

step-by-step explanation:

these have the exact same coordinates, thus they are the correct function.

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