Answer
Only $ACB $ is defined
$ACB=\left[\begin{array}{llll}
-3 & -21 & 27 & -6\\
-2 & -14 & 18 & -4
\end{array}\right]$
Work Step by Step
If $A$ is an $m\times n$ matrix and $B$ is an $n\times k$ matrix
(so the number of columns of $A$ is the same as the number of rows of $B$),
then the matrix product $AB$ is the $m\times k$ matrix
whose $ij$-entry is the inner product of the $i\mathrm{t}\mathrm{h}$ row of $A$ and the jth column of $B.$
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$ABC:$
$AB$ is not defined, B does not have 2 rows, as A has columns.
$ABC$ is not defined
$ACB:$
$AC$ is defined ( a $2\times 4$ matrix multiplies a $4\times 1$ matrix) and is a 2$\times$1 matrix
$ACB=(AC)B$
is defined ( a $2\times 1$ matrix multiplies a $1\times 4$ matrix) and is a $2\times 4$ matrix
$BAC$
$BA$ is not defined, A does not have 4 rows, as B has columns.
$BAC$ is not defined
$BCA:$
$CA$ is not defined, A does not have 4 rows, as $C$ has columns.
$BCA$ is not defined
$CAB $is not defined because $AB$ is not defined.
$CBA $is not defined because $BA$ is not defined.
Only $ACB $ is defined
$AC=\left[\begin{array}{llll}
1 & 0 & 6 & -1\\
2 & 1/2 & 4 & 0
\end{array}\right]\left[\begin{array}{l}
1\\
0\\
-1\\
-2
\end{array}\right]=\left[\begin{array}{l}
1(1)+0(0)+6(-1)+(-1)(-2)\\
2(1)+\frac{1}{2}(0)+4(-1)+0(-2)
\end{array}\right]=\left[\begin{array}{l}
-3\\
-2
\end{array}\right]$
$(AC)B=\left[\begin{array}{l}
-3\\
-2
\end{array}\right][1\quad \quad 7 \quad-9 \quad 2]$
$=\left[\begin{array}{llll}
-3 & -21 & 27 & -6\\
-2 & -14 & 18 & -4
\end{array}\right]$
