In a group of 115 men, 80 are professors and 72 are pastors. If 3 men are neither professors nor pastors, how many are both professors and pastors, and how many are pastors only?

Here's how to solve the problem using set theory:

Let $U$ be the total group of men.
Let $P$ be the set of Professors.
Let $A$ be the set of Pastors.

Given information:
Total number of men, $n(U) = 115$.
Number of Professors, $n(P) = 80$.
Number of Pastors, $n(A) = 72$.
Number of men who are neither Professors nor Pastors, $n(P' \cap A') = 3$.

Step 1: Find the number of men who are either Professors or Pastors (or both).
The total number of men is the sum of those who are in at least one category and those who are in neither category.

$$n(U) = n(P \cup A) + n(P' \cap A')$$ $$115 = n(P \cup A) + 3$$ $$n(P \cup A) = 115 - 3$$ $$n(P \cup A) = 112$$

i) Find the number of men who are Both Professors and Pastors.
We use the Principle of Inclusion-Exclusion for two sets:

$$n(P \cup A) = n(P) + n(A) - n(P \cap A)$$

Substitute the known values:

$$112 = 80 + 72 - n(P \cap A)$$ $$112 = 152 - n(P \cap A)$$ $$n(P \cap A) = 152 - 112$$ $$n(P \cap A) = 40$$

The number of men who are both Professors and Pastors is $\boxed{\text{40}}$.

ii) Find the number of men who are Professors nor Pastors.
This refers to the men who are neither Professors nor Pastors. This value is given directly in the problem statement.

$$n(P' \cap A') = 3$$

The number of men who are neither Professors nor Pastors is $\boxed{\text{3}}$.

iii) Find the number of men who are Pastors only.
The number of men who are Pastors only is the total number of Pastors minus those who are also Professors.

$$n(A \text{ only}) = n(A) - n(P \cap A)$$ $$n(A \text{ only}) = 72 - 40$$ $$n(A \text{ only}) = 32$$

The number of men who are Pastors only is $\boxed{\text{32}}$.