Consider the statements:
$\textbf{(1)}\ p\text{ }\wedge\sim q\wedge r\qquad\textbf{(2)}\ \sim p\text{ }\wedge\sim q\wedge r\qquad\textbf{(3)}\ p\text{ }\wedge\sim q\text{ }\wedge\sim r\qquad\textbf{(4)}\ \sim p\text{ }\wedge q\text{ }\wedge r$
where $p,q$, and $r$ are propositions. How many of these imply the truth of $(p\rightarrow q)\rightarrow r$?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$
Statement $1$ states that $p$ is true and $q$ is false.  Therefore, $p \rightarrow q$ is false, because a premise being true and a conclusion being false is, itself, false.  This means that $(p \rightarrow q) \rightarrow X$, where $X$ is any logical statement (or series of logical statements) must be true - if your premise is false, then the implication is automatically true.  So statement $1$ implies the truth of the given statement.
Statement $3$ similarly has $p$ as true and $q$ is false, so it also implies the truth of the given statement.
Statement $2$ states that $p$ and $q$ are both false.  This in turn means that $p \rightarrow q$ is true.  Since $r$ is also true from statement $2$, this means that $(p \rightarrow q) \rightarrow r$ is true, since $T \rightarrow T$ is $T$.  Thus statement $2$ implies the truth of the given statement.
Statement $4$ states that $p$ is false and $q$ is true.  In this case, $p \rightarrow q$ is true - your conclusion can be true even if your premise is false.  And, since $r$ is also true from statement $4$, this means $(p \rightarrow q) \rightarrow r$ is true.  Thus, statement $4$ implies the truth of the given statement.
All four statements imply the truth of the given statement, so the answer is $\boxed{\textbf{(E)}}$