List 4 partitions of the alphabet containing 3 sets. How many different partitions can be made? The first part of the question is easy; there are many right answers but I put down:
$$(a), (b), (c..z)$$
$$(a,b), (c), (d..z)$$
$$(a,b,c), (d), (e..z)$$
$$(a,b,c,d), (e), (f..z)$$
The part I am not getting at is the second part where it asks how many different partitions can be made? My friend hinted at using the principle of inclusion-exclusion but I do not know how to use it here? 
 A: Although this is not specified, it looks as if you  want to count the partitions of the alphabet into three non-empty subsets.  
We first solve a different problem, finding the number of ways to divide the alphabet into three non-empty teams, to be labelled Team $1$, Team $2$, and Team $3$.
So we want to count the number of surjective (onto) functions from a set $A$ of $26$ elements to the set $\{1,2,3\}$.
There are $3^{26}$ functions from $A$ to $\{1,2,3\}$. But some functions are not surjective. We count the non-surjective ones. 
There are $2^{26}$ functions from $A$ to the set $\{1,\2\}$, also $2^{26}$ to the set $\{2,3\}$, also $2^{26}$ to the set $\{1,3\}$. 
But if we add these three numbers, we will double-count the $3$ constant functions. So there are $3\cdot 2^{26}-3$ non-surjective functions, and therefore $3^{26}-3\cdot 2^{26}+3$ surjective functions.
Now take off the labels. Any (unlabelled) partition comes from $3!$ divisions into labelled teams. The number of partitions into $3$ sets is therefore 
$$\frac{3^{26}-3\cdot 2^{26}+3}{3!}.$$   
