Combinatorics question on a $3\times 3$ array The letters $AAABBBCCD$ are to be arranged in a $3\times 3$ array in such a way that every (horizontal) row contains three different letters and one of the diagonals contain three copies of a single letter.  
How many arrangements are possible?  
This is what I tried:
There are $2$ ways of choosing a diagonal.
There are $2$ choices of letters ($A$ or $B$) for this diagonal
Then there are $\binom{3}2$ ways of distributing the $C$'s in different rows.
With the letter we did not choose in step $2$, we take those and distribute it in $3$ rows in $\binom{3}2$ ways.
The $D$ now only has $1$ choice to go.  
Total is $2\times 2\times 3 \times 3 = 36$ ways.
Is this correct flow of logic?
 A: You overlooked how many ways the Cs can be placed in the rows you select for them.
Method 1:  Choose a diagonal in two ways.  Choose which letter, A or B, fills that diagonal.  Take the other letter that appears three times.  Choose one of the two open positions in each of the three rows for that letter (which guarantees we will have three different letters in each row).  Choose which row will receive the D.  The remaining positions must be filled with Cs.  Hence, there are 
$$2 \cdot 2 \cdot 2^3 \cdot 3 = 96$$
admissible arrangements.
Method 2: We modify your approach.
Choose a diagonal in two ways.  Choose which letter, A or B, fills that diagonal. Choose two of the three rows for the Cs.  In each selected row, choose which of the two open positions receives a C.  We must place a B in each of the three rows.  In the row with two open positions, choose which will be filled with a D.  The remaining three positions must be filled with Bs.
$$2 \cdot 2 \binom{3}{2} \cdot 2^2 \cdot 2 = 96$$
Method 3: We solve the problem the alternative way you considered in the comments.
Choose a diagonal in two ways.  Choose which letter, A or B, fills that diagonal.  Choose a row for D.  Choose one of the two open positions in that row for D.  The other must be filled with a B.  That leaves two rows with two open positions.  Each of those rows must be filled with a B and a C.  Therefore, each such row can be filled in two ways.
$$2 \cdot 2 \cdot 3 \cdot 2 \cdot 2!^2 = 96$$
A: My humble try:
Assume one of A or B is chosen for the diagonal in 2 number of ways. The diagonals are 2 in number. Just for now assume you have selected A  for the diagonal, Then the first  row's selection could be 
BC, CD, BD 
The second row selection is dependent on the first row
if BC, again a BC or BD 
if BD, again a BC
BC-BC-BD
BC-BD-BC
BD-BC-BC
Each one can be permuted in 2 ways and there are 3 ways this could happen.
Thus
The total number of ways $= 2.2.2^3.3 = 96$.
