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$e^+e^-\rightarrow e^+e^-$ is called Bhabha scattering. Let us only consider the tree level Feynman diagrams of this process. Apparantly, there are s-channel and t-channel diagrams as shown in the linked wiki page enter link description here. But there should also be a u-channel diagram obtained by modifying the "Scattering" diagram:

enter image description here

Sorry for bad hand drawing.

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You only need to include $u$ and $t$ channels when the final particles are indistinguishable (classically thinking, you don't know which blip in your detector came from which initial particle). Since $e^-$ and $e^+$ are distinguishable, you will always know which one is which and only $t$ channel contributes to the scattering amplitude.

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