Count the total number of subsets such that $\bigcup\limits_{i=1}^{3} T_{i} = P$ Given a set $P = \{1,2,3\}$. We define another set $S = P \times P \times P$. Count the number of tuples $(T_1,T_2,T_3)$, where $T_i$ is a subset of the set $P$, such that $\bigcup\limits_{i=1}^{3} T_{i} = P$.
My approach - My guess is that I have to apply principle of inclusion and exclusion here. If I count the total number of such tuples that can be formed using the set and then remove the unwanted cases, like the subsets that don't contain a single element from the set $P$, then I'll get the answer. However, I'm having trouble counting the number of tuples in each case.
Is my approach correct? Any help is appreciated.
 A: Another approach is to consider which sets the elements of $P$ lie in. There are $3$ elements in $P$. For $\cup_{i=1}^3 T_i=P$, we require that each element $k\in P$ is in at least one of $T_i$.
Each $T_i$ has $2$ choices: contain $k$ or do not contain $k$. This gives $2^3$ possibilities. However, we cannot have that all $T_i$ choose to not contain $k$, hence there are $\boxed{2^3-1}$ possibilities for each $k\in P$.
Since there are $3$ different values of $k$ that we need to consider, and each of these $k\in P$ have $2^3-1$ ways to be placed in the sets $T_i$, there are $\boxed{\left(2^3-1\right)^3=343}$ ordered triples $(T_1,T_2,T_3)$

For fun, let's generalize the problem so that $|P|=n$ and we want to find the number of ordered $k$-tuples $(T_1,T_2,\ldots T_k)$.
Using the same reasoning, for any element $e\in P$, we have $2^k-1$ ways for it to appear in the $k$-tuple $(T_1,T_2,\ldots T_k)$ so that it appears at least once.
Since we have to account for each of these $n$ elements, the total number of ordered $k$-tuples is $\left(2^k-1\right)^n$. In your problem, we have $n,k=3$, which yields $\left(2^3-1\right)^3=\boxed{343}$
