# 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?

Sep 21, 2017

$259459200$

#### 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. $\frac{30}{2} = 15$

Then we have 15 choices distributed over the 8 questions.

Using the formula for permutations:

(n!)/((n - r)!)

Where $n$ is the number of objects ( In this case the marks in groups of 2).

And $r$ is how many are taken at a time ( In this case the 8 questions)

So we have:

(15!)/(( 15 - 8)!) = (15!)/(7!) = 259459200

Sep 23, 2017

There are ""_21C_14 (or 116,280) ways.

#### Explanation:

We start with 30 marks in the "bank" to give. Since all questions must be worth at least 2 marks, we take $2 \times 8 = 16$ marks from the $30$ and distribute them equally. Now each question has 2 (so far) and the "bank" is left with $30 - 16 = 14$ marks.

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 15!), then dividing by the number of ways to permute both yellow marbles (14!) and blue marbles (1!), since within each arrangement, it doesn't matter in which order the identical marbles appear.

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 ""_15C_14.

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! ways to arrange all marbles, then divide out by the ways to permute both the yellow ones (14!) and the blue ones (2!):

(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 ""_16C_14.

By now, you may notice where we're headed. The number to the left of the $C$ is equal to the number of marks we're splitting (yellow marbles) 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 $C$ stays the number of marks.

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

${\text{_(14+7)C_14=}}_{21} {C}_{14}$

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.