Find the number of triplets (A,B,C)? Find the number of triplets $(A,B,C)$ where A,B,C are subsets of $\{1,2,3...8\}$ such that $ A \cap B \cap C = \phi $, $ A \cap B \neq \phi $, $ B \cap  C \neq \phi$.
Edit: My approach, split the space into $A, B, C,  A \cap B $, $ B \cap C $, and universal. So each element can be placed at $6$ places that is $ 6^8 $ now you cannot  have the $ A \cap B $ and $ B \cap C $ empty so subtract the case when they are empty. That is $ 2\cdot 5^8 $ . Which doesn't seem right.
 A: Your idea was OK, but some corrections are needed . . .

There are $7$ mutually exclusive locations for an element to be placed, namely

*

*$A$ only,$\;B$ only,$\;C$ only$\\[4pt]$

*$A\cap B,\;B\cap C,\;C\cap A$$\\[4pt]$

*$(A\cup B\cup C)\,'$
By hypothesis, we have $A\cap B\cap C={\large{\varnothing}}$ which allows us to consider $7$ locations rather than $8$. It's also what allows us to claim that the three pairwise intersections are mutually exclusive.
Applying the principle of inclusion-exclusion
$$7^8-2{\,\cdot\,}6^8+5^8=2796194$$
is the count we seek.

Explanation:

*

*$7^8$ counts all assignments of the $8$ elements to the $7$ locations, temporarily ignoring the restrictions on $A\cap B$ and $B\cap C$.$\\[4pt]$

*To correct the overcount, we subtract the $6^8$ assignments for which $A\cap B={\large{\varnothing}}$, and we also subtract the $6^8$ assignments for which $B\cap C={\large{\varnothing}}$, which explains the subtraction of $2{\,\cdot\,}6^8$.$\\[4pt]$

*But now we have to add back the $5^8$ assignments for which both $A\cap B={\large{\varnothing}}$ and $B\cap C={\large{\varnothing}}$, since those assignments were subtracted twice in the previous step, and they should only be subtracted once.

