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Hey Vishal,

That’s an excellent question that was also asked during the lecture.

But to reiterate, here is another possible argument you could have used:

Without regard to order, we split the $20$ students into 2 groups $A$ and $B$. If both Olivia and Oliver are in group A, then the remaining $18$ students can be arranged into $2$ groups in $\dbinom{18}{10}$ ways since $10$ of them must go to group $B$ and $8$ of them must go to group A randomly. Similarly, we also have $\dbinom{18}{10}$ arrangements if both Oliver and Olivia are in group B. Thus, our desired solution is $\frac{2\dbinom{18}{10}}{\dbinom{20}{10}}=\frac{9}{19}$.