Use the Inclusion-exclusion principle in order to count the number of positive integers $\le 1000$ that can't be divided by $7, 11$ and $13$. 
Use the Inclusion-exclusion principle in order to count the number of positive integers $\le 1000$ that can't be divided by $7, 11$ and   $13$.

This excercise was given to us, but I feel like it doesn't make much sense. In the past, we had similar excercises that would, for example, ask for the number of positive integers $\le 1000$ that can't be divided by $7, 11$ or $13$. This kind of question can be directly answered with the Inclusion-exclusion principle, but the excercise above actually has two variables in it, which are $|A_7 \cup A_{11} \cup A_{13}|$ (number of positive integers that can't be divided by $7, 11$ or $13$) and $|A_7 \cap A_{11} \cap A_{13}|$ (number of positive integers that can't be divided by $7, 11$ and $13$), and in order to get the first one, I would have to calculate the second one, and this would already be the answer to the question above, without applying the Inclusion-exclusion principle at all.
Is there another way to do this that I am missing here?

If it's "and":
$$
1000-\left\lfloor\frac{1000}{1001}\right\rfloor=1000
$$
If it's "or":
$$
\scriptsize1000-\left\lfloor\frac{1000}7\right\rfloor-\left\lfloor\frac{1000}{11}\right\rfloor-\left\lfloor\frac{1000}{13}\right\rfloor+\left\lfloor\frac{1000}{77}\right\rfloor+\left\lfloor\frac{1000}{91}\right\rfloor+\left\lfloor\frac{1000}{143}\right\rfloor-\left\lfloor\frac{1000}{1001}\right\rfloor=720
$$
