This is the statement we need to translate into predicate logic:
Every professor bores some of his/her students.
$P(x)$: $x$ is a professor.
$S(x,y)$: $x$ is a student of $y$.
$B(x,y)$: $x$ bores $y$.
I came up with
$$\forall x \exists y \left( P(x) \land S (y,x) \to B(x,y) \right)$$
but the answer in the solutions is
$$\forall x \exists y \left( P(x) \to S (y,x) \land B(x,y) \right)$$
Why is my answer incorrect?
Remember than an implication $p \to q$ evaluates to True whenever $p$ is False. In your answer, you only need to find a $y$ such that $P(x) \land S(y,x)$ is False for your implication $P(x) \land S(y,x) \to B(x,y)$ to evaluate to True.
In English, your implication would be satisfied if, for every professor, there were a person who was not a student of that professor (which seems like a reasonable assertion!), even if every professor had exactly zero bored students.