Find number of skew-symmetric matrices of order $3\times 3$ in which all non-diagonal elements are different Find number of skew-symmetric matrices of order $3\times 3$ in which all non-diagonal elements are different and belong to the set $\{-9,-8,-7,...,7,8,9\}$
My Attempt:
I did a simple calculation and obtained $$\binom{9}{3}\times(3!)\times 2^3=4032$$
But answer given is $$\frac{4032}{6}=672$$
Why has it been divided by $6$
 A: The given answer would be correct if the problem specified "up to permutations of rows and columns". But it did not, and your answer is correct.
A: As suggested by @aarbee in comment section i am providing a detailed solution to given problem.
Definition: $\;(i)\;$Since it is skew symmetric matrix so all the elements in diagonal will be zero.
$\;(ii)\;$ $a_{ij}=−a_{ji}\;∀\;i≠j$ for skew symmetric matrix.
Explanation of $9\choose 3$
First choose $3$ elements either from set $\{−9,−8,−7,−6,−5,−4,−3,−2,−1\}$ or $\{1,2,3,4,5,6,7,8,9\}$ because $a_{ij}=−a_{ji}\;∀\;i≠j$ for skew symmetric matrix. So number of ways to choose $3$ elements out of $9$ is ${9\choose 3}$
and these three entries can be arranged in three place say $a_{12},a_{13},a_{23}$ in $3!$ ways
We can't do ${18 \choose 3}\cdot3!$ because in this way we can get $3$ elements $1,-1,\;2$ which won't form skew-symmetric matrix because in this case $3$ elements in upper diagonal or lower diagonal will $1 , -1, \; 2$.
And $2^3$ because we can permute $a_{12}$ and $a_{21},\;$ $\;a_{13}$ and $a_{31},\;$ $\;a_{23}$ and $a_{32}$.
So  total number of ways are  ${9\choose 3}\cdot3!\cdot2^3$
Detailed explanation of $2^3$: Let say selected elements are $1,2,3$
Here are those $8$ matrcies whose elmenets are $1,2,3$
$\begin{bmatrix} 
 0 & 1 & 2\\
 -1 & 0 & 3\\ -2 & -3 & 0\\
 \end{bmatrix},\;$ $\begin{bmatrix} 
 0 & -1 & 2\\
 1 & 0 & 3\\ -2 & -3 & 0\\
 \end{bmatrix},\;$ $\begin{bmatrix} 
 0 & 1 & -2\\
 -1 & 0 & 3\\ 2 & -3 & 0\\
 \end{bmatrix},\;$ $\begin{bmatrix} 
 0 & 1 & 2\\
 -1 & 0 & -3\\ -2 & 3 & 0\\
 \end{bmatrix},\;$ $\begin{bmatrix} 
 0 & -1 & -2\\
 1 & 0 & 3\\ 2 & -3 & 0\\
 \end{bmatrix},\;$ $\begin{bmatrix} 
 0 & -1 & 2\\
 1 & 0 & -3\\ -2 & 3 & 0\\
 \end{bmatrix},\;$ $\begin{bmatrix} 
 0 & 1 & -2\\
 -1 & 0 & -3\\ 2 & 3 & 0\\
 \end{bmatrix},\;$ $\begin{bmatrix} 
 0 & -1 & -2\\
 1 & 0 & -3\\ 2 & 3 & 0\\
 \end{bmatrix},\;$.
