= Solution
Number the <simple roots> of the $A_4$ <root system> consecutively along its <Dynkin diagram>. For a <Young diagram> with row lengths $r_i$, its <highest weight> has <Dynkin labels> $a_i=r_i-r_{i+1}$. The three <irreducible representations> have the following <Young diagrams>, which can be filled increasingly down the column to give <Young tableaux>:
$$
\begin{array}{c|c|c|c}
&\text{column of boxes}&\text{tensor}&\text{highest-weight labels}\\\hline
\mathbf5&\boxed{1}&v^i&(1,0,0,0)\\
\overline{\mathbf5}&\begin{matrix}\boxed{1}\\\boxed{2}\\\boxed{3}\\\boxed{4}\end{matrix}
&w^{[i_1i_2i_3i_4]}\ \text{or }\widetilde v_i&(0,0,0,1)\\
\mathbf{10}&\begin{matrix}\boxed{1}\\\boxed{2}\end{matrix}
&T^{ij}=-T^{ji}&(0,1,0,0)
\end{array}
$$
Here $i=1,\ldots,5$, $v^i\mapsto U^i{}_jv^j$, and $T^{ij}\mapsto U^i{}_kU^j{}_lT^{kl}$. Contraction with the invariant alternating <tensor> identifies $\widetilde v_i=(1/4!)\epsilon_{ii_1i_2i_3i_4}w^{i_1i_2i_3i_4}$, giving the <dual representation> of $\mathbf5$. The dimensions follow from $\binom51=5$, $\binom54=5$, $\binom52=10$.
Use the <centralizer> and <circle group> normalization of <3/d/solution>. The first three components have <U(1) charge> $-2$ and the last two have <U(1) charge> $3$, so the <branching of a representation> is
$$
\boxed{\mathbf5\longrightarrow(\mathbf3,\mathbf1)_{-2}\oplus(\mathbf1,\mathbf2)_3,\qquad
\overline{\mathbf5}\longrightarrow(\overline{\mathbf3},\mathbf1)_2\oplus(\mathbf1,\mathbf2)_{-3}.}
$$
For the <dual representation>, both <U(1) charges> reverse and the $SU(3)$ <representation> is conjugated; the $SU(2)$ doublet is equivalent to its conjugate. Taking the <exterior square> of the first decomposition gives three types of antisymmetric <tensor> components: $T^{ab}$, $T^{a\alpha}$, and $T^{\alpha\beta}$, where $a,b=1,2,3$ and $\alpha,\beta=4,5$. Their <U(1) charges> add, and $\Lambda^2\mathbf3=\overline{\mathbf3}$, $\Lambda^2\mathbf2=\mathbf1$. Hence
$$
\boxed{\mathbf{10}\longrightarrow
(\overline{\mathbf3},\mathbf1)_{-4}\oplus(\mathbf3,\mathbf2)_1\oplus(\mathbf1,\mathbf1)_6.}
$$
The dimensions are $3+6+1=10$. These product labels describe genuine <representations> of the quotient <group> in <3/d/solution>: every element of its $\mathbb Z_6$ <kernel> acts trivially on each summand.
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