Find the number of $4 \times 4$ orthogonal matrices whose entries are either $0$ or $1$

Question

Find the number of $4 \times 4$ orthogonal matrices whose entries are either $0$ or $1$.

My solution

It is the same as the "permutation" of the four orthonormal columns
$$\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0\\ \end{pmatrix}\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0\\ \end{pmatrix}\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0\\ \end{pmatrix}\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1\\ \end{pmatrix}.$$

So the answer is $4!=24$. Could you please check my solution?

Answer 1

You are right. Perhaps that you coud have added that, since the matrix is orthogonal, no column can have only null entries and, on the other hand, if one of the entries is $1$, then all other must be equal to $0$.