Assume we have 8 elements arranged in two rows (A, B, C, D, a, b, c, d)
it is known that the number of arrangements is 8! = 40320
What I thought about is:
Step 1: We assume the two rows initially have the elements
$\begin{bmatrix} A&B&C&D \\ a&b&c&d \end{bmatrix}$
Step 2: We first have arrange both rows
so this allows
$\begin{bmatrix} A&B&C&D \\ b&a&c&d \end{bmatrix}$, $\begin{bmatrix} A&B&C&D \\ b&a&c&d \end{bmatrix}$, $\begin{bmatrix} D&C&B&A \\ b&a&c&d \end{bmatrix}$, $\begin{bmatrix} D&C&B&A \\ d&a&c&b \end{bmatrix}$ etc.
$\therefore(4!)^2=576$
Step 3: We then arrange the elements in each columns
so this allows (using above examples)
$\begin{bmatrix} b&B&C&d \\ A&a&c&D \end{bmatrix}$, $\begin{bmatrix} b&a&c&D \\ A&B&C&d \end{bmatrix}$, $\begin{bmatrix} b&a&c&d \\ D&C&B&A \end{bmatrix}$, $\begin{bmatrix} D&C&c&A \\ d&a&B&b \end{bmatrix}$ etc.
$\therefore(2!)^4=16$
Step 4: Calculate the number of arrangements: $(4!)^2\times(2!)^4=9216$
Unfortunately, this answer is obviously different from $8!$, but I couldn't observe the mistake in this approach. Any help will be much appreciated, thanks in advance.
Your approach ensures that each column has exactly 1 upper case and 1 lower case letter, which is not true in general.