# The number of way in which a examiner can assign 30 marks to 8 questions given not less than 2 marks to any question is?

Then, there are 15 options spread across the eight questions.

Applying the permutation formula:

Thus, we have:

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There are

The only thing left to do is figure out how many ways there are to divide the final 14 marks between the 8 questions. This may sound very difficult at first, but there is a trick that makes it much easier.

Let's simplify for a moment. If there were only two questions and 14 marks were to be divided between them, how many different ways could we divide the marks? We could divide the marks as 14 + 0, or 13 + 1, or 12 + 2, etc.... or 1 + 13, or 0 + 14. To put it another way, there are 15 different ways to introduce a single split (between two questions).

Consequently, if there are one blue marble (question splitter) and fourteen yellow marbles (marks), there are

Therefore, in order to distribute the final 14 marks among the eight questions (requiring seven splitters), we compute

Hence, there are 116,280 ways to divide the 30 marks among the 8 questions, with a minimum of 2 marks assigned to each question.

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To find the number of ways an examiner can assign 30 marks to 8 questions, given that each question must receive at least 2 marks:

First, allocate 2 marks to each of the 8 questions, which would account for (2 \times 8 = 16) marks.

Then, the remaining marks to be distributed is (30 - 16 = 14).

Since each question must receive at least 2 marks, and there are 8 questions, it means there are (8 - 1 = 7) remaining "slots" for marks to be distributed.

This problem can be approached using a stars and bars method. We can visualize this by representing each mark as a star, and the bars as dividers between the questions. Since there are 14 marks left to distribute, there are (14 + 7 = 21) total objects (stars + bars).

The number of ways to arrange these objects can be calculated using the combination formula, (C(n + r - 1, r)), where (n) is the total number of objects (21 in this case), and (r) is the number of bars (7 in this case).

So, the number of ways the examiner can assign 30 marks to 8 questions, given not less than 2 marks to any question, is (C(21, 7)).

Calculating (C(21, 7)) yields:

[ C(21, 7) = \frac{21!}{7!(21 - 7)!} = \frac{21!}{7!14!} = 11628 ]

Therefore, there are 11628 ways for the examiner to assign the marks to the 8 questions under the given conditions.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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