An interesting combinatorics problem $A$ is a set containing $n$ elements. A subset $P_1$ of $A$ is chosen. The set is reconstructed by replacing the elements of $P_1$. Then a subset $P_2$ is chosen and again the set is reconstructed by replacing the elements of $P_2$. In this way $m$ subsets $P_1,......,P_m$ are chosen where $m>1$.
Find the number of ways of choosing $P_1,.......,P_m$ such that no two of them  are pairwise disjoint.
I don't have any clue how to start this problem
I tried a lot of things but couldn't even get close to a conclusion.

Edit: An equivalent (and hopefully simpler and neater) way of putting it.
Given a set $A$ with $n$ elements, let $B=(P_1,P_2 \cdots, P_m)$ be a tuple of $m$ non-empty subsets of $A$, $P_i \subset A$ such that $P_i \cap P_j \ne \varnothing$. How many different $B$ there are?
 A: For the simplicity I will draw it for m=3 and generalize it along the way by putting it into brackets.
You have to put n elements into $2^3=8$ areas ($2^m$ areas), as on the image bellow.

Each goes in one of them. Total ways to do that is $8^n$ ($2^{mn}$). By doing it you define one triplet $P_1,P_2,P_3$ [m-plet $P_1,...,P_m$]
But you have to exclude those distributions of elements when some of the lemons remain empty. Lemons are areas $P_i\bigcap P_j$, I will call them L(i,j).
Similar I will define $L(i_1,...,i_k)=P_{i_1}\cap...\cap P_{i_k}$. Number of simple areas $L(i_1,...,i_k)$ covers is $2^{m-k}$ since it may or may not  intersect with m-k other subsets.
To count all the distributions of elements when some of the lemons L(i,j) remain empty can be done by inclusion-exclusion principle.
Let $C(L(i_1,j_1),...,L(i_q,j_q))$ be count of all the distributions of the elements such that each of the lemons remain empty.
We can see that $C(L(i,j))=(2^m-2^{m-2})^n$, $i<j$. From total number of simple ares we subtract lemons number of simple areas...
But for $q>1$ it gets a bit complicated. Let see for $q=2$. It may be $C(L(i,j),L(i,k))$ or $C(L(i,j),L(k,l))$ (notice that for second case m has to be greater than 3)
Number of areas that $L(i,j),L(i,k)$ cover is not same as the number $L(i,j),L(k,l)$ cover.
$C_{2,a}=C(L(i,j),L(i,k))=(2^m-2\cdot 2^{m-2}+2^{m-3})^n$
$C_{2,b}=C(L(i,j),L(k,l))=(2^m-2\cdot 2^{m-2}+2^{m-4})^n$
Lets see distributions when 3 lemons are excluded. It may be:
$C_{3,a}=C(L(i_1,i_2),L(i_1,i_3),L(i_2,i_3))=(2^m-3\cdot 2^{m-2}+3\cdot 2^{m-3}-2^{m-3})^n$
$C_{3,b}=C(L(i_1,i_2),L(i_1,i_3),L(i_1,i_4))=(2^m-3\cdot 2^{m-2}+3\cdot 2^{m-3}-2^{m-4})^n$
$C_{3,c}=C(L(i_1,i_2),L(i_1,i_3),L(i_2,i_4))=(2^m-3\cdot 2^{m-2}+2\cdot 2^{m-3}+2^{m-4}-2^{m-4})^n$
$C_{3,d}=C(L(i_1,i_2),L(i_1,i_3),L(i_4,i_5))=(2^m-3\cdot 2^{m-2}+2^{m-3}+2\cdot 2^{m-4}-2^{m-5})^n$
$C_{3,e}=C(L(i_1,i_2),L(i_3,i_4),L(i_5,i_6))=(2^m-3\cdot 2^{m-2}+3\cdot 2^{m-4}-2^{m-6})^n$
In the end it will be:
$2^{mn}-{m\choose2}C(L(i,j))+{m\choose3}{3\choose1}C_{2,a}+{m\choose4}{4\choose2}C_{2,b}-{m\choose3}C_{3,a}-{m\choose4}{4\choose1}C_{3,b}-{m\choose4}{4\choose1}{3\choose1}C_{3,c}-{m\choose5}{5\choose1}{4\choose2}C_{3,d}-{m\choose6}{6\choose2}{4\choose2}C_{3,e}+...$
Maybe now you find some law how this formulas are generated...
A: The first step to solving such a problem is to define precisely what we are trying to answer.
For all $n>0,m>0$, let us define $C(n,m)$ as the number of distinct tuples $(P_1, P_2, \cdots P_m)$ where each $P_j$ (with $1\le j\le m$) is a set of natural numbers less than or equal to $n$ (in symbols, $\forall e\in P_j\mid 1\le e\le n$) and no two sets within the same tuple are disjoint; that is, for any tuple being counted $\neg\exists(j,k)\mid\{P_j\cap P_k=\emptyset\}$.
Your question is to compute $C(n,m)$.
First let us note that $C(1,m)=1$ for all $m$.
Likewise, let us note that $C(n,1)=2^n-1$ for all $n$.
(To be continued.)
