According to Wikipedia, a Quaternion can be represent in 48 different $4\times4$ matrix forms. Finding this number can be proved by using permutation?
$$\binom 4 2 \cdot4=\frac{4!}{(4-2)!}\cdot4=48$$
As in each column, there are $\begin{pmatrix} 4 \\ 2 \end{pmatrix}$ different permutations. Is this correct?
For example, two of those 48 representations are mentioned below:
\begin{align} & \begin{bmatrix}a&-b&-c&-d\\b&a&-d&c\\c&d&a&-b\\d&-c&b&a\end{bmatrix} \\[10pt] = {} & a \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix} +b \begin{bmatrix}0&-1&0&0\\1&0&0&0\\0&0&0&-1\\0&0&1&0\end{bmatrix} +c \begin{bmatrix}0&0&-1&0\\0&0&0&1\\1&0&0&0\\0&-1&0&0\end{bmatrix} +d \begin{bmatrix}0&0&0&-1\\0&0&-1&0\\0&1&0&0\\1&0&0&0\end{bmatrix} \end{align}
OR \begin{align} & \begin{bmatrix} a & b & c & d \\ -b & a & -d & c \\ -c & d & a & -b \\ -d & -c & b & a \end{bmatrix} \\[10pt] = {} & a\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}+ b\begin{bmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{bmatrix}+ c\begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{bmatrix}+ d\begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{bmatrix} \end{align}
But my main question is how they can be found? Obviously, the reason is because of Hamilton's famous equation:
$$\mathbf{i}^2 =\mathbf {j}^2=\mathbf {k}^2 = \mathbf {i} \mathbf {j} \mathbf {k} =-1$$
But how?