= Solution
Take $n=6$ in 6(b). The <symplectic quotient of the binary subset module> has dimension four, so the action gives a <group homomorphism>
$$
S_6\longrightarrow Sp_4(2).
$$
It is injective. If $\sigma$ acts trivially on the quotient, every two-element subset $A$ has $\sigma A+A\in\{\varnothing,\Omega\}$. Hence $\sigma A$ equals $A$ or its complement. The latter has size four, whereas $\sigma A$ has size two; therefore $\sigma$ fixes every two-element subset. Given $i$, choose distinct $j,k\ne i$. It then fixes the intersection $\{i,j\}\cap\{i,k\}=\{i\}$, so it fixes each point and is the identity.
Finally 6(a) gives
$$
|Sp_4(2)|=2^4(2^2-1)(2^4-1)=16\cdot3\cdot15=720=|S_6|.
$$
An injective <group homomorphism> between these <finite groups> of the same order is surjective. Thus
$$
\boxed{Sp_4(2)\cong S_6.}
$$
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