To solve this problem, we need to determine the number of ways to form a committee of 8 students from the council, such that there are exactly two representatives from each class (freshmen, sophomores, juniors, and seniors).

Step-by-Step Solution

1. Choose 2 representatives from the 3 freshmen: $\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3 \times 2}{2 \times 1} = 3$

2. Choose 2 representatives from the 4 sophomores: $\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6$

3. Choose 2 representatives from the 3 juniors: $\binom{3}{2} = 3$

4. Choose 2 representatives from the 5 seniors: $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$

5. Calculate the total number of committees: Since the selections are independent, multiply the number of combinations for each group: $\text{Total combinations} = \binom{3}{2} \times \binom{4}{2} \times \binom{3}{2} \times \binom{5}{2} = 3 \times 6 \times 3 \times 10 = 540$

However, the choices given are 220, 48, 1960, and 22, so let's recheck the problem setup:

If the requirement was for 2 representatives from each class and no other conditions were given, the solution as above is correct. But the actual question setup does not match the result 540, which suggests something may have been overlooked.

Given the provided choices and the typical distribution of such problems, the correct approach is indeed as calculated. But if the answer was supposed to be among those listed, then there might be a contextual error in the problem.

The most plausible correct answer (based on our method) would be None of the above choices since the correct number of ways is 540.

But as none of the choices given match this, if forced to choose, you should check the problem statement with your instructor.