= Solution
The <symmetric square of a direct sum> gives
$$
\operatorname{Sym}^2(A\oplus B)=\operatorname{Sym}^2A\oplus(A\otimes B)\oplus\operatorname{Sym}^2B.
$$
Now $\operatorname{Sym}^2\mathbf2=\mathbf3$ and $\operatorname{Sym}^2\mathbf3=\mathbf6$, rather than the singlet and antitriplet arising from <exterior squares>. Thus
$$
\boxed{\mathbf{15}\downarrow=(\mathbf3,\mathbf1)_6\oplus(\mathbf2,\mathbf3)_1\oplus(\mathbf1,\mathbf6)_{-4}.}
$$
The dimensions are $3+6+6=15$.
By contrast, <3/c/solution> gives the following left-handed <Weyl fermion> content of $\overline{\mathbf5}\oplus\mathbf{10}$:
$$
L:(\mathbf2,\mathbf1)_{-3},\quad d^c:(\mathbf1,\overline{\mathbf3})_2,\quad
Q:(\mathbf2,\mathbf3)_1,\quad u^c:(\mathbf1,\overline{\mathbf3})_{-4},\quad
e^c:(\mathbf1,\mathbf1)_6.
$$
These are one <Standard Model> family, with <weak hypercharge> $Y=q_c/6$. The $\mathbf{15}$ contains $Q$ but lacks $L$, $d^c$, $u^c$ and $e^c$; instead it contains an <SU(2)> triplet and an <SU(3)> sextet. Equality of total dimensions therefore does not give the same <representation> content, and $\mathbf{15}$ cannot supply the family.
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