In my opinion, all combinations make sense,
Implication:
$\forall{x}(P \to C)$: For all $x$, if $P$ is true then $C$ is also true.
$\exists{x}(P \to C)$: There exists atleast a $x$ such that if $P$ is true then $C$ is also true.
Conjunction:
$\forall{x}(P \wedge C)$: For all $x$, Both $P$ and $C$ are true.
$\exists{x}(P \wedge C)$: There exists atleast a $x$, such that both $P$ and $C$ are true.
However, if you will post specific example where you find problem/confusion, question will be more clear.
