In how many ways can two rocks, one red and one blue and each covering $2$ cells, be placed on a chess board? A chess board has 8 rows and 8 columns.
We take two, 2-cells size rock (it covers 2 cells) one is blue and one is red.
In how many ways we can put these stones on the board?
I first calculated how many ways I can put one stone which is $7\times8\times2$ (for vertical and horizontal assignment).
I think that I should next calculate the number of potential places for the second stone, but that varies if the first stone is horizontal or vertical, or if it has an odd or even number of cells away from its corners.
Can someone tell me how to calculate this correctly?
 A: The number of places for the second stone indeed depend on the placement of the first one. So we can calculate the number of places for the 2nd stone conditionally on the 1st stone. 
It is in fact easier to calculate the number of place where you can not place the second stone.
Let suppose also the first stone is placed horizontally as we will used symmetry to get the total number. We have 56 cases placement for the first stone.
Let $n_2(i,j)$ be the number of place where you can not place the second stone.
(i,j) represents the position of the left cell covered by first stone, starting by the left up corner. So for instance (1,1) is the left up corner and (8,7) means the right down corner.
We have then:
Corners cases
$n_2(1,1) = 4$. By symmetry $n_2(1,7) = n_2(8,1) = n_2(8,7) =4$ (4 cases)
First and last row cases
$n_2(i,j) = 5$ for $i \in \{1,8\}$ and $j \in [2,6]$ (10 cases)
First and last columns cases
$n_2(i,j) = 6$ for $i \in [2,7]$ and $j \in \{1,7\}$ (12 cases)
Middle cases
$n_2(i,j) = 7$ for $i \in [2,7]$ and $j \in [2,6]$ (30 cases)
We have dealt with $30+12+10+4=56$ cases so with all the possibilities of horizontal placement for the first stone.
Remember we have a number of $7*8*2=112$ placements if we are not counting the overlap.
Now we sum all the cases:
$(112-4)*4+(112-5)*10+(112-6)*12+(112-7)*30 = 5924$.
$5924$ is the number of cases when the first stone is placed horizontally.If it is placed vertically , it is the same number as the board is completely symmetric ( just rotate the chess board by 90° and the first stone will be horizontal:)
Therefore the total number is $5924*2=11848$
