= Solution
Let $p_1,p_2$ be the incoming and $p_3,p_4$ the outgoing <four-momenta>, and set $s=(p_1+p_2)^2$, $t=(p_1-p_3)^2$, $u=(p_1-p_4)^2$, the <Mandelstam variables>. The only connected <tree-level Feynman diagram> is the $s$-channel exchange of $\bar\Phi$, with <Noether charge> $+2$. The two incoming light <particles> attach to $\phi^2\Phi$, and the two outgoing light <particles> to its conjugate. There is no four-light contact <Feynman vertex>, nor a charge-allowed $t$ or $u$ exchange with these external labels.
\Image[/past-exam-of-the-quantum-field-theory-course-of-the-university-of-oxford/2025-light-scalar-scattering.png]
{title=Light-particle scattering through a heavy antiparticle in the s channel}
Define the <invariant scattering amplitude> by $\langle f|S-1|i\rangle=i(2\pi)^4\delta^{(4)}(P_f-P_i)\mathcal M$. Then
$$
i\mathcal M=(-ih)\frac{i}{s-M^2+i0}(-ih),\qquad
\boxed{\mathcal M(\phi\phi\to\phi\phi)=-\frac{h^2}{s-M^2+i0}.}
$$
The <Feynman vertices> already contain the identical-leg combinatorics. No extra factor of two multiplies this <invariant scattering amplitude>; the distinct <identical final-state symmetry factor> enters a <cross section> integral.
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