Count the number of multiples of $3,\,5,\,7,\,11,\,13$ in the first $1000$ numbers I know it can perfectly be done with the inclusion-exclusion principle but I think it will be boring to count the cardinality of the intersections and because they're less than a "big" number, 1000. Is there an easier way to count them?
 A: There are floor(1000/3) multiples of 3.  4/5 of them are not multiples of 5.  So there are floor(floor(100/3)4/5) multiples of 3 that are not multiples of 5.  By same reasoning the are floor (12/13 floor(10/11 floor (6/7 floor (4/5 floor (1000/3))))) multiples of 3 that are not multiples of 5,7,11,or13.  Call this M3.
Likewise there are floor(12/13 floor (10/11 floor (6/7 floor (1000/5)))) multiples of 5 (some of which are multiples of 3 the we omitted before) that are not multiples of 7,11,13.  Call this M5.
Do the same for M7, M11,and M13.  
Add these up.  This counts the multiples of lower numbers without double counting the mutual multiples of higher numbers.  Then we "catch" up the mutual multiples when we do the higher numbers.
I'm not sure how much tedium this eliviates.
A: For reference if we have a set $P$  of primes and we want to count the
multiples of the elements  of $P$ less than some number  $N$ we get by
inclusion-exclusion for the non-multiples
$$\sum_{S\subseteq P} 
(-1)^{|S|} \bigg\lfloor \frac{N}{\prod_{p \in S} p}\bigg\rfloor.$$
Introducing $M$ as the product of the primes in $P$ this becomes
$$\sum_{d|M} \mu(d) 
\bigg\lfloor \frac{N}{d}\bigg\rfloor.$$
This because  the Moebius  function from number  theory is  the Mobius
function of the  divisor poset. Subtract these values from  $N$ to get
the number of multiples.
Doing the computation for $P=\{3,5,7,11,13\}$ and $N=1000$ we get
$$N-\sum_{S\subseteq P} 
(-1)^{|P|} \bigg\lfloor \frac{N}{\prod_{p \in S} p}\bigg\rfloor
= 618.$$
In case we are willing to accept an approximation we obtain
$$\sum_{d|M} \mu(d) 
\bigg\lfloor \frac{N}{d}\bigg\rfloor
\approx N \sum_{d|M} \frac{\mu(d)}{d}
= N \prod_{p|M} \left(1-\frac{1}{p}\right)
\\ = N \times\frac{1}{M}\times M \prod_{p|M} \left(1-\frac{1}{p}\right)
= N \frac{\varphi(M)}{M}.$$
This agrees with what we would expect from inspection and it yields
for the present problem
$$1000 - 1000 \times \frac{5760}{15015} \approx
616.3836164$$
which is an acceptable approximation.

Remark. Supposing that  we do not have  a set $P$ of  primes but a
set $Q$  of positive integers  which are not necessarily  co-prime the
formula becomes
$$\sum_{S\subseteq Q} 
(-1)^{|S|} \bigg\lfloor \frac{N}{\mathrm{lcm}(S)}\bigg\rfloor.$$
