Can you make an unlimited number of connectives with $p$ and $q$? One of the questions in my exercise asks:

We have made several logical connectives using the propositions p and
  q. For instance p ∧ q, p ∨ q and p ↑ q . Can this go on indefinitely;
  that is, making an unlimited number of connectives with p and q ? Or
  is there a finite number of distinct connectives? If so, can you
  determine how many there can be?

I was trying to come up with an answer but got stuck. Does it mean an unlimited number of connectives with p and q such that 
$$p \land q$$ 
$$p \land (p \land q)$$
$$p \land (p \land (p \land q))$$
In that way you definitely can keep going forever, or for another example:
$$ \neg p $$
$$ \neg (\neg p)$$
$$ \neg (\neg (\neg p))$$
So does it mean 

Can you create infinite connectives using the same operation?

or 

How many connectives can be made using unique operations?

Also, for the last question 

If so, can you determine how many there can be?

Wouldn't it be the number of conjunctions available? (That mean  something different to each other)
I would really appreciate some help and a better explanation of the question at hand, maybe even hints so that I can complete it on my own.
Thank you in advance.

It's not talking about "infinite connectives," it's talking about an infinite number of connectives.
Two ostensibly different-looking connectives can actually "the same," for instance de Morgan's law tells us $\neg (p\wedge q)$ and $(\neg p)\vee(\neg q)$ are equivalent. Whether two are equivalent or not can be determined by looking at the truth table.
How many possible truth tables are there involving $p$, $q$ and a third column?
