To solve this problem, we will consider the different ways in which the student can select questions from two distinct sections under the given constraint that at least 2 questions must be selected from each section. We can break it down into different cases and compute the number of possibilities for each case.

There are three cases to consider:

Case 1: Selecting 2 questions from Section I and 4 questions from Section II.
Case 2: Selecting 3 questions from Section I and 3 questions from Section II.
Case 3: Selecting 4 questions from Section I and 2 questions from Section II.

Then we will add the number of ways from all the cases, as these are mutually exclusive events.

Let's calculate for each case:

Case 1: Choose 2 questions from Section I and 4 questions from Section II.

The number of ways to choose 2 questions from 6 in Section I is given by $C(6, 2)$, and the number of ways to choose 4 questions from 5 in Section II is given by $C(5, 4)$.

Therefore, the total number of ways for Case 1 = $C(6, 2) \times C(5, 4)$.

Case 2: Choose 3 questions from Section I and 3 questions from Section II.

The number of ways to choose 3 questions from 6 in Section I is given by $C(6, 3)$, and the number of ways to choose 3 questions from 5 in Section II is given by $C(5, 3)$.

Therefore, the total number of ways for Case 2 = $C(6, 3) \times C(5, 3)$.

Case 3: Choose 4 questions from Section I and 2 questions from Section II.

The number of ways to choose 4 questions from 6 in Section I is given by $C(6, 4)$, and the number of ways to choose 2 questions from 5 in Section II is given by $C(5, 2)$.

Therefore, the total number of ways for Case 3 = $C(6, 4) \times C(5, 2)$.

Let's calculate the number of combinations for each case:

For Case 1: $C(6, 2) \times C(5, 4) = \frac{6!}{2!(6-2)!} \times \frac{5!}{4!(5-4)!} = 15 \times 5 = 75$
For Case 2: $C(6, 3) \times C(5, 3) = \frac{6!}{3!(6-3)!} \times \frac{5!}{3!(5-3)!} = 20 \times 10 = 200$
For Case 3: $C(6, 4) \times C(5, 2) = \frac{6!}{4!(6-4)!} \times \frac{5!}{2!(5-2)!} = 15 \times 10 = 150$

Now to find the total number of different ways, we add the number from all three cases together:

Total number of ways = $75 + 200 + 150 = 425$

Thus, the total number of different ways a student can select 6 questions, taking at least 2 questions from each section, is 425, which corresponds to Option A.

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