For the grammar below, a partial LL(1) parsing table is also presented along with the grammar. Entries that need to be filled are indicated as E1, E2, and E3. $\varepsilon$ is the empty string, $\$$ indicates end of input, and $\mid$ separates alternate right hand sides of productions. $S \rightarrow aAbB \mid bAaB \mid \varepsilon$ $A \rightarrow S$ $B \rightarrow S$ The appropriate entries for E1, E2, and E3 are

To find the entries of the LL(1) parsing table $M[A, a]$:

1. For non-terminal $S$:

   • Production $S \rightarrow aAbB$: $\text{FIRST}(aAbB) = \{a\}$, so add $S \rightarrow aAbB$ to $M[S, a]$.

   • Production $S \rightarrow bAaB$: $\text{FIRST}(bAaB) = \{b\}$, so add $S \rightarrow bAaB$ to $M[S, b]$.

   • Production $S \rightarrow \varepsilon$: since $\varepsilon \in \text{FIRST}(\varepsilon)$, add $S \rightarrow \varepsilon$ to $M[S, t]$ for all t \in \text{FOLLOW}(S) = \{a, b, \\}$, i.e.,$M[S, a]$,$M[S, b]$, and$M[S, \$]$.

   Therefore:

   • E1$= M[S, a] = \{S \rightarrow aAbB,\ S \rightarrow \varepsilon\}$

   • E2$= M[S, b] = \{S \rightarrow bAaB,\ S \rightarrow \varepsilon\}$

2. For non-terminal $B$:

   • Production $B \rightarrow S$: $\text{FIRST}(S) = \{a, b, \varepsilon\}$.

   • Add $B \rightarrow S$ to $M[B, a]$ and $M[B, b]$.

   • Since $S \Rightarrow^* \varepsilon$, add $B \rightarrow S$ to $M[B, t]$ for all t \in \text{FOLLOW}(B) = \{a, b, \\}$, which includes$M[B, \$]$.

   Therefore:

   • E3 = M[B, \] = \{B \rightarrow S\}$

This matches option (C).