= Symmetric traceless SU(3) tensor representation
{title2=$d(p,q)=\tfrac12(p+1)(q+1)(p+q+2)$}
Symmetrize the $p$ upper and $q$ lower indices separately, and remove every upper-lower <trace>. The resulting representation has <highest weight> $p\omega_1+q\omega_2$. Its <dimension> is
$$
\binom{p+2}{2}\binom{q+2}{2}-\binom{p+1}{2}\binom{q+1}{2}
=\frac{(p+1)(q+1)(p+q+2)}2.
$$
The subtraction is valid because contraction is onto: in the polynomial realization its adjoint, for the monomial-factorial <inner product>, is multiplication by the nonzero polynomial $\sum_ix_iy_i$, an injective map. The resulting kernel contains the highest-weight constituent and has precisely its <Weyl dimension formula> <dimension>, hence is irreducible. This highest-weight constituent is not always the largest-dimensional constituent of the full mixed <tensor>.
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