# 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?

##### 2 Answers

#### Explanation:

If I am reading this correctly, then if the examiner can assign marks only in multiples of 2. This then would mean there are only 15 choices out of the 30 marks .i.e.

Then we have 15 choices distributed over the 8 questions.

Using the formula for permutations:

Where

And

So we have:

There are

#### Explanation:

We start with 30 marks in the "bank" to give. Since all questions must be worth at least 2 marks, we take

Now we just need to find the number of ways to divvy up the remaining 14 marks among the 8 questions. At first, this may seem very hard, but there's a trick that makes it much more intuitive.

Let's simplify things for a moment. What if we only had 2 questions, and 14 marks to split between them? How many ways could we do that? Well, we could split the marks as 14 + 0, or 13 + 1, or 12 + 2, etc. ... or 1 + 13, or 0 + 14. In other words, when we only need to introduce 1 split (between 2 questions), we get 15 ways to do it.

This is the same as asking, "How many unique ways can we arrange 14 yellow marbles (the marks) and 1 blue marble (the question splitter) in a row?" The answer to this is found by calculating the number of permutations of all 15 marbles (which is

So when there are 14 yellow marbles (marks) and 1 blue marble (question splitter), there are

#(15!)/(14!xx1!)=(15xxcancel(14!))/(cancel(14!)xx1)=15/1=15#

15 ways to arrange the marbles (split the marks). Note: this is equal to

Let's introduce another blue marbleâ€”that is, a second split, or a third question to give the marks to. Now we have 16 total marbles, and we want to know how many unique ways we can arrange these. Similar to before, we take the

#(16!)/(14!xx2!)=(16xx15xxcancel(14!))/(cancel(14!)xx2xx1)=(16xx15)/(2)=120#

So there are 120 ways to split 14 marks between 3 questions. This is also equal to

By now, you may notice where we're headed. The number to the left of the **plus** the number of splitters (blue marbles). The number of splitters is always *one less than* the number of questions. The number to the right of the

Thus, to split the remaining 14 marks among all 8 questions (which requires 7 splitters), we calculate

#""_(14+7)C_14=""_21C_14#

#color(white)(""_(14+7)C_14)=(21!)/(7!xx14!)#

#color(white)(""_(14+7)C_14)="116,280"#

So there are 116,280 ways to assign 30 marks to 8 questions, where each question is worth at least 2 marks.