In order to analyse the relationship between $E_m, L_m$ and $S_m$ it is convenient to have a closer look at the sets which form the building blocks for these numbers.
In case of problem (a) it boils down to show that according to the formula of $E_m$
\begin{align*}
\color{blue}{E_2}&=\sum_{j=2}^4(-1)^{j-2}\binom{j}{2}S_j\\
&\,\,\color{blue}{=S_2-\binom{3}{1}S_3}+\color{blue}{\binom{4}{2}S_4}\tag{1}
\end{align*}
and we also want to clarify how to derive the binomial coefficients $\binom{3}{1}$ and $\binom{4}{2}$.
The Setting:
Given the word ARRANGEMENT, we consider the set
\begin{align*}
X=\{\mathrm{AAEEGMNNRRT},\ldots,\mathrm{ARRANGEMENT},\ldots,\mathrm{TRRNNMGEEAA}\}
\end{align*}
consisting of $\frac{11!}{\left(2!\right)^4}=2\,494\,800$ permutations with repetitions of the letters from this word.
We introduce a set $U$ of properties
\begin{align*}
U=\{P_A,P_E,P_N,P_R\}
\end{align*}
A word in $X$ has property $P_A\in U$ if the letter A occurs consecutively and the meaning of the other properties in $U$ is analogously.
Given a set of $T\subseteq U$ of properties from $U$ we define
$E(T)$ as the number of words in $X$ which have exactly the properties $T\subseteq U$, and
$L(T)$ as the number of words in $X$ which have at least the properties $T\subseteq U$.
So, for instance $\mathrm{ARRANGEMENT}$ contributes to $E(\{P_R\})$ and $L(\{P_R\})$, whereas $\mathrm{AAEEGMNNRRT}$ contributes to $L(\{P_R\})$ and $U$, but not to $E(\{P_R\})$.
The numbers $E(T)$ and $L(T)$ form building blocks for $E_m$ and $L_m$. Since we have
$E_m$ is the number of words which have exactly $m$ properties from $U, (0\leq m\leq 4)$ and
$L_m$ is the number of words which have at least $m$ properties from $U, (0\leq m\leq 4)$
we can write these quantities as
\begin{align*}
\color{blue}{E_m}&\color{blue}{=\sum_{{T\subseteq U}\atop{|T|=m}}E(T)}\tag{2.1}\\
\color{blue}{L_m}&\color{blue}{=\sum_{{T\subseteq U}\atop{|T|\geq m}}E(T)}\tag{2.2}\\
\end{align*}
Example $m=2$:
In the case $m=2$ we have according to (2.1)
\begin{align*}
E_2&=\sum_{{T\subseteq U}\atop{|T|=2}}E(T)\\
&=E(\{P_A,P_E\})+E(\{P_A,P_N\})+E(\{P_A,P_R\})\\
&\qquad + E(\{P_E,P_N\})+E(\{P_E,P_R\})+E(\{P_N,P_R\})\\
&=\binom{4}{2}E(\{P_A,P_E\})
\end{align*}
where the factor $\binom{4}{2}=6$ can be taken thanks to the symmetry of the words with its respective properties. For $L_2$ we obtain according to (2.2)
\begin{align*}
L_2&=\sum_{{T\subseteq U}\atop{|T|\geq 2}}E(T)
=\sum_{j=2}^{4}\sum_{{T\subseteq U}\atop{|T|=j}}E(T)\\
&=E_2+E_3+E_4\tag{2.3}
\end{align*}
The Setting continued:
Now we take a closer look at the quantities $S_m$. They are calculated for $0\leq m\leq 4$ as
\begin{align*}
S_m=\frac{(11-2\times m+m)!}{(2!)^{4-m}}\binom{4}{m}\tag{2.4}
\end{align*}
and their meaning is:
- $S_m$ is the number of words in $X$, so that for each subset $T\subseteq U$ of properties with size $|T|=m$ we take the number of words which have at least $m$ properties from $T\subseteq U$, i.e.
\begin{align*}
\color{blue}{S_m=\sum_{{T\subseteq U}\atop{|T|=m}}L(T)}\tag{2.5}
\end{align*}
Indeed, taking for instance $m=2$ in (2.4) we have
\begin{align*}
S_2=\frac{(11-2\times 2+2)!}{(2!)^{4-2}}\binom{4}{2}
\end{align*}
where the factor $\binom{4}{2}$ represents each subset $T\subseteq U$ of properties with size $|T|=2$, whereas the factor $(11-2\times 2+2)!$ in the numerator represents the number of permutations of the other letters besides the selected two pairs. Since there are two more letters which occur twice we have to divide by $2^{4-2}=4$ as they can't be distinguished.
$S_m$ is given in (2.5) in terms of $L(T)$. We can find a more convenient representation in terms of $E(T)$ as follows. We have in case $m=2$:
\begin{align*}
\color{blue}{S_2}=\color{blue}{\sum_{{T\subseteq U}\atop{|T|=2}}L(T)}
&=\sum_{{T\subseteq R\subseteq U}\atop{|T|=2}}E(R)\\
&=\sum_{R\subseteq U}E(R)\sum_{{T\subseteq R}\atop{|T|=2}}1\\
&=\sum_{R\subseteq U}E(R)\binom{|R|}{2}\\
&=\sum_{j=2}^4\sum_{{R\subseteq U}\atop{|R|=j}}E(R)\binom{j}{2}\\
&=\sum_{j=2}^4\binom{j}{2}E_j\\
&\,\,\color{blue}{=E_2+\binom{3}{2}E_3+\binom{4}{2}E_4}\tag{2.6}
\end{align*}
A derivation of (2.6) in a more general context is given in this answer.
Answer of (a) and (b):
We find in the same way as in (2.4) to following identities:
\begin{align*}
S_2&=\sum_{j=2}^4\binom{j}{2}E_j=E_2+\binom{3}{2}E_3+\binom{4}{2}E_4\tag{3.1}\\
S_3&=\sum_{j=3}^4\binom{j}{3}E_j=E_3+\binom{4}{3}E_4\tag{3.2}\\
S_4&=\sum_{j=4}^4\binom{j}{4}E_j=E_4\tag{3.3}
\end{align*}
These relations enable us to represent $E_2$ in terms of $S_2, S_3$ and $S_4$. We obtain
\begin{align*}
\color{blue}{E_2}&=S_2-\binom{3}{2}E_3-\binom{4}{2}E_4\tag{$\to (3.1)$}\\
&=S_2-\binom{3}{2}\left(S_3-\binom{4}{3}E_4\right)-\binom{4}{2}E_4\tag{$\to (3.2)$}\\
&=S_2-\binom{3}{2}S_3+\left(\binom{3}{2}\binom{4}{3}-\binom{4}{2}\right)S_4\tag{$\to (3.3)$}\\
&\,\,\color{blue}{=S_2-\binom{3}{2}S_3+\binom{4}{2}S_4}
\end{align*}
according to result (a). Similarly we obtain with (2.3)
\begin{align*}
\color{blue}{L_3}&=E_3+E_4\\
&=S_3-\binom{4}{3}E_4+E_4\tag{$\to (3.2)$}\\
&=S_3-\left(\binom{4}{3}-1\right)S_4\tag{$\to (3.3)$}\\
&\,\,\color{blue}{=S_3-\binom{3}{1}S_4}
\end{align*}
according to result (b).
