p B and 3 questions from group C, or 2 questions from group A, 3 questions from group B and 2 questions from group C, or 3 questions from group A, 2 questions from group B and 2 questions from group C. These are the only possible combinations.

Let's calculate the number of ways to choose each of these combinations:

2 from A, 2 from B, 3 from C:
The number of ways to choose 2 questions from group A is $${}^4C_2 = \frac{4!}{2!2!} = 6$$.

The number of ways to choose 2 questions from group B is $${}^5C_2 = \frac{5!}{2!3!} = 10$$.

The number of ways to choose 3 questions from group C is $${}^6C_3 = \frac{6!}{3!3!} = 20$$.

So, the total number of ways to choose this combination is 6 * 10 * 20 = 1200.
2 from A, 3 from B, 2 from C:
The number of ways to choose 2 questions from group A is $${}^4C_2 = 6$$.

The number of ways to choose 3 questions from group B is $${}^5C_3 = \frac{5!}{3!2!} = 10$$.

The number of ways to choose 2 questions from group C is $${}^6C_2 = \frac{6!}{2!4!} = 15$$.

So, the total number of ways to choose this combination is 6 * 10 * 15 = 900.
3 from A, 2 from B, 2 from C:
The number of ways to choose 3 questions from group A is $${}^4C_3 = \frac{4!}{3!1!} = 4$$.

The number of ways to choose 2 questions from group B is $${}^5C_2 = 10$$.

The number of ways to choose 2 questions from group C is $${}^6C_2 = 15$$.

So, the total number of ways to choose this combination is 4 * 10 * 15 = 600.

Therefore, the total number of ways the candidate can make his selection is 1200 + 900 + 600 = 2700.

So the answer is Option C.

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