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?

Answer 1

#259459200#

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. #30/2 = 15#

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

Applying the permutation formula:

#(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)

Thus, we have:

#(15!)/(( 15 - 8)!) = (15!)/(7!) = 259459200#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

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

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

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).

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.

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

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

#""_(14+7)C_14=""_21C_14#
#color(white)(""_(14+7)C_14)=(21!)/(7!xx14!)#
#color(white)(""_(14+7)C_14)="116,280"#

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.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 3

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.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

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.

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.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7