= Solution
Let $V$ have <dimension> $2m$ over $\mathbb F_q$, with a nondegenerate <alternating bilinear form> $B$. The symplectic <group> is
$$
Sp_{2m}(q)=\{g\in GL(V):B(gv,gw)=B(v,w)\}.
$$
It acts freely and transitively on ordered symplectic bases. Choose the first vector $e_1\ne0$ in $q^{2m}-1$ ways, then choose $f_1$ with $B(e_1,f_1)=1$ in $q^{2m-1}$ ways. Their span is nondegenerate, so choose the remaining <symplectic basis> in its <orthogonal complement>. This gives the recurrence
$$
|Sp_{2m}(q)|=(q^{2m}-1)q^{2m-1}|Sp_{2m-2}(q)|,
$$
and therefore
$$
\boxed{|Sp_{2m}(q)|=q^{m^2}\prod_{j=1}^m(q^{2j}-1),\qquad |Sp_4(2)|=2^4(2^2-1)(2^4-1)=720.}
$$
For the <permutation> module $\mathbb F_2^r$ with $r$ even, put $\mathbf t=(1,\ldots,1)$ and
$$
W=\left\{x:\sum_i x_i=0\right\}=\mathbf t^\perp.
$$
It is invariant under coordinate <permutations> and has codimension one. The <dot product> restricted to $W$ is alternating because $B(x,x)=\sum_i x_i^2=\sum_i x_i=0$. Since the ambient <dot product> is nondegenerate, $W^\perp=\langle\mathbf t\rangle$. Evenness of $r$ ensures $\mathbf t\in W$, so
$$
\boxed{\operatorname{rad}(B|_W)=\langle\mathbf t\rangle.}
$$
Here the word symplectic must be understood as an alternating form before quotienting: a form with a nonzero radical is degenerate. The nondegenerate <symplectic form> is induced on $\overline W=W/\langle\mathbf t\rangle$, of <dimension> $r-2$.
Now put $r=2m+2$. Coordinate <permutations> preserve the induced form and give a homomorphism $S_r\to Sp_{r-2}(2)$. To see it is faithful for $r\ge6$, suppose a <permutation> $\sigma$ acts trivially on $\overline W$. For every pair of distinct indices,
$$
e_{\sigma(i)}+e_{\sigma(j)}\equiv e_i+e_j\pmod{\langle\mathbf t\rangle}.
$$
The two weight-two vectors could differ by $\mathbf t$ only if the complement of a two-element set also had size two. For $r\ge6$ that is impossible. Thus $\sigma$ preserves every two-element set, and hence fixes every index: intersect two such sets with one common element. Therefore
$$
\boxed{S_{2m+2}\hookrightarrow Sp_{2m}(2)\quad(m\ge2).}
$$
For $m=2$, both <groups> have order $720$, and the injection is onto. This proves
$$
\boxed{S_6\cong Sp_4(2).}
$$
The lower bound matters: for $r=4$ the action on the quotient has the <Klein four-group> as <group homomorphism kernel>. The construction is the <deleted permutation module in characteristic two>.
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