The symmetry of clebsch-gordan coefficients $\left< j_1j_2;m_1m_2 \middle| j_1j_2;JM \right>$ under exchange of $j_1,m_1$ and $j_2,m_2$ is \begin{equation} \left< j_1j_2;m_1m_2 \middle| j_1j_2;JM \right>=(-1)^{j_1+j_2-J}\left< j_2j_1;m_2m_1 \middle| j_2j_1;JM \right> \end{equation} Now consider the case in which $j_1=2$, $j_2=1$, $J=2$, $m_1=m_2=1$ and $M=2$.
Applying the above formula to this we get \begin{equation} \left< 21;11 \middle| 21;22 \right>=-\left< 12;11 \middle| 12;22 \right> \end{equation} I think this should mean $\left< 21;11 \middle| 21;22 \right>=0$ as the order of $j_1$ and $j_2$ shouldn't matter and $\left< 21;11 \middle| 21;22 \right>$ and $\left< 12;11 \middle| 12;22 \right>$ must be the same state. But it is non-zero.
The same logic works for $m_1=m_2=0$ though and $\left< 21;00 \middle| 21;20 \right>=0$, but it might just be a coincidence. It seems I have not fully understood what the symmetry relation means. Why must there be a phase factor at all?
In many places the $j_1$ and $j_2$ are omitted and a blind application of the symmetry equation will give $\left<11 \middle| 22 \right>=-\left<11 \middle| 22 \right>$, and hence we might conclude that $\left<11 \middle| 22 \right>=0$.