= Solution
Write the defining space as $V=A\oplus B$, where $A\cong\mathbb C^2$ and $B\cong\mathbb C^3$. Use ordered pairs for the <SU(2)> and <SU(3)> factors, in that order. The generator $X_c$ in <3/b/solution> gives <U(1) charges> $3$ on $A$ and $-2$ on $B$, so
$$
\boxed{\mathbf5\downarrow=(\mathbf2,\mathbf1)_3\oplus(\mathbf1,\mathbf3)_{-2}.}
$$
Taking the <conjugate representation> reverses the <U(1) charges>. The <SU(2)> doublet is equivalent to its <dual representation>, while the <SU(3)> triplet becomes the antitriplet:
$$
\boxed{\overline{\mathbf5}\downarrow=(\mathbf2,\mathbf1)_{-3}\oplus(\mathbf1,\overline{\mathbf3})_2.}
$$
For the <two-index antisymmetric representation>, split the <antisymmetric tensor> $T^{IJ}=-T^{JI}$ into $T^{ab}$, $T^{a\mu}$ and $T^{\mu\nu}$, with $a,b=1,2$ and $\mu,\nu=3,4,5$. The <exterior square of a direct sum> gives
$$
\Lambda^2(A\oplus B)=\Lambda^2A\oplus(A\otimes B)\oplus\Lambda^2B.
$$
The first term is an <SU(2)> singlet, the middle is $(\mathbf2,\mathbf3)$, and $\Lambda^2\mathbf3\cong\overline{\mathbf3}$ by contraction with the invariant alternating <tensor>. Adding the <U(1) charges> of the two indices gives
$$
\boxed{\mathbf{10}\downarrow=(\mathbf1,\mathbf1)_6\oplus(\mathbf2,\mathbf3)_1\oplus(\mathbf1,\overline{\mathbf3})_{-4}.}
$$
The dimensions $1+6+3=10$ check the <branching of a representation>. In the convention $Q_{\rm em}=T_3+Y$, <weak hypercharge> is $Y=X_c/6$.
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