How many $k$ tuples of subsets $S_1,...,S_k$ are there such that $S_1 \cap ..... \cap S_k = \emptyset $. How many $k$ tuples of subsets $S_1,...,S_k$ are there such that 
$S_1 \cap ..... \cap S_k = \emptyset  $.
An attempt: Suppose we can make a sketch of a Venn diagram with $k$ subsets. Then there are $2^k$ total locations in the diagram, but we don't have to take their intersection since it has to be empty. Thus there are $2^{k} -1$ places to place $n$ numbers from $[n] = (1,...,n)$. Then there are $n-2^k \choose 2^k - 1$ total ways of doing this.
Can someone please provide some feedback and tell me if I am in the right track ?
I would really appreciate it. Thank you.
 A: Throughout, let $[n] = \{1, 2, \cdots, n\}$, and let $\mathcal{P}(S)$ denote the power set of $S$.
Given a $k$-tuple $(S_1, \cdots, S_k)$ of subsets of $[n]$ such that $\bigcap_{1\le i \le k} S_i = \emptyset$, consider the function $f: [n] \to \mathcal{P}([k])$ that takes $x \in [n]$ to the set of indices $i$ such that $x \in S_i$. In particular, since $\bigcap S_i = \emptyset$, we have $f(x) \neq [k]$ for all $x \in [n]$.
This function $f$ totally determines the $k$-tuple, since $S_i = \{x \in [n] \, : \, i \in f(x)\}$. Conversely, given any function $f: [n] \to \mathcal{P}([k])$ with $f(x) \neq [k]$ for all $x \in [n]$, we can obtain such a $k$-tuple by setting $S_i = \{x \in [n] \, : \, i \in f(x)\}$, as before.
Thus, to count the number of $k$-tuples, it suffices to count the functions satisfying the condition, of which there are evidently $(2^k -1)^n$.
A: With this problem it does seem  to have its origins in a combinatorics
course   and  hence   we   may   suppose  that   it   is  asking   for
inclusion-exclusion. We get the formula (follows by inspection)
$$\sum_{q=0}^n {n\choose q} (-1)^q (2^{n-q})^k.$$
With  this  inclusion-exclusion  poset  the nodes  $P  \subseteq  [n]$
represent $k$-tuples whose  intersection is $P$ or a  superset of $P.$
This formula  assigns weight  one to $k$-tuples  who do  not intersect
because they only  appear in the $q=0$ term. Now for  a $k$-tuple that
intersects in  exactly $p$ elements it  is included in  all nodes that
are subsets of those $p$ elements. With $q$ the size of these nodes we
get the contribution
$$\sum_{q=0}^p {p\choose q} (-1)^q = 0$$
because $p\ge  1.$ Therefore we  assign weight one only  to admissible
tuples  and  weight   zero  to  all  others,  which   is  the  desired
behaviour. 
Simplifying the formula we get
$$2^{nk} \sum_{q=0}^n {n\choose q} (-1)^q 2^{-kq}
= 2^{nk} (1-2^{-k})^n = (2^k-1)^n$$
which matches the result from the accepted answer.
A: You may also compute the number of such $k-$ tuples by considering the complementary set.
For a fixed $X\subseteq A$ with $X\neq \emptyset$  we have $$|\{A_1 \cap \dots \cap A_k =X\}|=(2^{n-|X|})^k$$
Moreover, $$\big\{A_1 \cap \dots \cap A_k \neq \emptyset\big\}=\bigcup_{X\subseteq A \\ X\neq \emptyset}\{A_1 \cap \dots \cap A_k = X\}$$ Since the above union is disjoint, $$\begin{eqnarray*}|\big\{A_1 \cap \dots \cap A_k \neq \emptyset\big\}| &=& \sum_{X\subseteq A \\ X \neq \emptyset } |\{A_1 \cap \dots \cap A_k = X\}| \\ &=& \sum_{j=1}^n \sum_{X\subseteq A \\ |X|=j} |\{A_1 \cap \dots \cap A_k = X\}| \\ &=& \sum_{j=1}^n \sum_{X\subseteq A \\ |X|=j} (2^{n-j})^k \\ &=& \sum_{j=1}^n {n \choose j}(2^{k})^{n-j} \\ &=& \sum_{j=0}^n{n \choose j}(2^{k})^{n-j} - 2^{nk}   \\ &=& (1+2^{k})^n-2^{nk} \end{eqnarray*}$$ So, $$|\{A_1 \cap \dots \cap A_k\ = \emptyset\}|=2^{nk}-|\{A_1 \cap \dots \cap A_k \neq \emptyset\}|=2^{nk+1}-(2^k+1)^n$$
