First, count all the ways you can do it when the top row is all orange. Then each additional row must have exactly two oranges. There are $6$ ways of choosing two columns for each of these rows, but you cannot pick the same pair of columns for each row, and you cannot pick two rows with exact complements. If you pair a set with its omplement, then that means you need to pick one row from each of the pairs.
The pairs are:
$\{\{1,2\},\{3,4\}\}$ $\{\{1,3\},\{2,4\}\}$ $\{\{1,4\},\{2,3\}\}$
So, $\{1,2\}$ corresponds to a row with orange on the columns $1$ and $2$.
There are $8$ ways of picking a row from each pair (and it turns out you can do so arbitrarily, although you have to show that,) and then $6$ ways to order them, so the total, when the top row is all orange, is $48.$
The more general solution is $48*16$, because you show that, for any top row, you get the same number of possible corresponding rows.
If you want to treat solutions the same if you can re-order the rows in one to get the other, then you have $32=48*16/24$ solutions.