The number of ways in which an examiner can assign 30 marks to 8 questions, giving not less than 2 marks to any question, is :

To solve this problem, we need to distribute 30 marks to 8 distinct questions such that each question gets at least 2 marks. We will use the method of distributing identical items with constraints.

First, since each question must have at least 2 marks, we assign 2 marks to each of the 8 questions. This uses up 2 × 8 = 16 marks. The total marks left to distribute are 30 - 16 = 14 marks.

Now, we have 14 identical marks to distribute freely among the 8 questions, with no restrictions (meaning a question can receive zero or more additional marks). This is a problem of distributing identical items into distinct groups.

The formula for the number of ways to distribute n identical items into k distinct groups is given by the binomial coefficient:

$$\binom{n + k - 1}{k - 1} \quad \text{or equivalently} \quad \binom{n + k - 1}{n}$$

Here, n = 14 (remaining marks) and k = 8 (questions). So, substituting the values:

$$\binom{14 + 8 - 1}{14} = \binom{21}{14}$$

Alternatively, we can write:

$$\binom{14 + 8 - 1}{8 - 1} = \binom{21}{7}$$

Since binomial coefficients are symmetric, we know that:

$$\binom{21}{14} = \binom{21}{7}$$

Therefore, the number of ways is $$\binom{21}{7}$$.

Now, comparing with the options:

A. $$^{30}C_7$$

B. $$^{21}C_8$$

C. $$^{21}C_7$$

D. $$^{30}C_8$$

Option C matches our result.

Hence, the correct answer is Option C.