Combinatorics - inclusion exclusion principle i have some question and i'm not sure i solved it well.
How much numbers between 1 to 800 (Naturals) are devided at least by one of the following numbers: 3,4,6.
This is my solution:
Total devided by 3: 800/3 = 266
Total devided by 4: 800/4 = 200
Total devided by 6: 800/6 = 133
Total devided by 3*4 = 800/12 = 66
Total devided by 3*6 = 800/18 = 44
Total devided by 4*6 = 800/24 = 33
Total devided by 3* 4*6 = 800/72 = 11
Total Answer = 266+200+133-66-44-33+11=467
Am I right?
Thanks!
 A: I agree with the other responses.  Further, the OP specifically asked
about using Inclusion-Exclusion against this problem.  Therefore, if
the other responses had not already answered the OP's specific question,
I would not have posted this response.
The problem can easily (and validly) be attacked without Inclusion-Exclusion.
To see this, first see this link, including its addendum.
Number of positive integers $\le n$ which are a multiple of $p$
Using the ideas from the link, you can attack the problem as follows:

*

*The problem is to identify all numbers that are multiples of 3, 4, or 6
from the set $\{1,2, \cdots, 800\}.$


*The number $6$ can be discarded from the investigation because any number that is a multiple
of $6$ must also be a multiple of $3$.  Therefore, the computation is the
same if you restrict the computation to all numbers that are multiples of
$3$ or $4$.


*$3$ and $4$ are relatively prime.  Further, $792$ is a multiple of
$3 \times 4$.


*This means that with attention temporarily restricted to the subset
$\{1,2,\cdots,792\}$, divisibility by $3$ and divisibility by $4$ can
be regarded as independent events.


*This means that in this subset, the chance of a number not being
divisible by either $3$ or $4$ is
$$\frac{4-1}{4} \times \frac{3-1}{3} = \frac{2}{4} = \frac{1}{2}.$$

*

*This means that in the set $\{1, 2, \cdots, 792\},$ exactly
$\frac{1}{2} \times 792 = 396$ numbers will be multiples of either
$3$ or $4$.


*The next step is to check the numbers $\{793, 794, \cdots, 800\}$
manually.


*In this small subset, the numbers $796$ and $800$ are divisible by
$4$ and the numbers $795$ and $798$ are divisible by $3$.


*This means that the final computation is $396 + 2 + 2 = 400.$
Addendum 
It occurs to me, after the fact, that the approach that I am suggesting should be formally proven.
Consider the numbers $k, m, n \in \mathbb{Z^+}$, where $m,n$ are relatively prime.
Let the set $S \equiv \{1, 2, \cdots, (k\times m \times n)\}.$
(Ironically), using Exclusion-Inclusion, the number of elements in $S$ that are either multiples of $m$ or multiples of $n$ are
$$(kn) + (km) - k.$$
This means that the number of elements in $S$ that are not multiples of
either $m$ or $n$ is
$$kmn - kn - km + k = k(mn - n - m + 1) = kmn \times
\frac{mn - n - m + 1}{mn} = kmn \times \frac{n-1}{n} \times \frac{m-1}{m}.$$
A: You need to look a bit closer at your intersections. The numbers that are divisible by both $3$ and $6$ are not just the numbers that are divisible by $18$. They are actually all the numbers that are divisible by $6$.
Similar mistakes happen with $4$ and $6$ (should be $12$, not $24$) and $3$, $4$ and $6$ (should be $12$, not $72$).
So the final answer is
$$
266+200+133-133-66-66+66=400
$$
