I am studying "A Baby Problem" (Ch. I.7) in Anthony Zee: Quantum Field Theory in a Nutshell. The generating functional for the $\Phi^4$ theory is evaluated to be

$$ \tilde{Z} \equiv \frac{Z[J, \lambda]}{Z[0,0]} = \exp(-\frac{\lambda}{4!}(\frac{\delta}{\delta J})^4) \exp(\frac{1}{2m^2}J^2) \,.$$

We can series expand both the exponentials, and obtain any term in the double series expansion of $\tilde{Z}$ in $\lambda$ and $J$.

Suppose we want the term of order $\lambda$ and $J^5$ in $\tilde{Z}$. We can extract the order $J^{10}$ term in $e^{(J^2/2m^2)}$, namely $[1/5!(2m^2)^5]J^{10}$, replace $e^{(-\frac{\lambda}{4!}(\frac{\delta}{\delta J})^4)}$ by $-\frac{\lambda}{4!}(\frac{\delta}{\delta J})^4$, and differentiate to get $ \sim [(-\lambda)/(m^2)^5]J^6$.

In a Feynman diagram, this term should have five propagators, six external legs and one vertex.

How is that possible?