How Many Ways Can You Choose Questions on a History Exam with Restrictions?

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Homework Help Overview

The problem involves a history exam with eight questions divided into two parts, A and B, each containing four questions. Candidates must select five questions, ensuring that at least two are from part A and at least two from part B. Participants are exploring the various ways to calculate the combinations under these restrictions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to apply combinatorial formulas to determine the number of ways to select questions while adhering to the given constraints. There are discussions about different combinations and the validity of certain approaches, with some questioning the assumptions behind the calculations.

Discussion Status

The discussion is ongoing, with various interpretations of the problem being explored. Some participants have offered insights into the structure of the problem, while others express confusion about the requirements and the implications of selecting questions from each part. There is no clear consensus yet on the correct approach.

Contextual Notes

Participants note that the problem's wording may lead to different interpretations regarding whether candidates must attempt exactly five questions or at least five. This ambiguity is contributing to the complexity of the discussion.

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Homework Statement


A history exam paper contains eight questions, four in part A and four in B. Candidates are required to attempt five questions. In how many ways can this be done if at least two questions from part and at least two questions from B must be attempted


Homework Equations





The Attempt at a Solution


I tried with (4 2) * (4 2) * (4 1) where (n r) is the no. of combinations of r objects from n objects. But it is the wrong answer. Give me the right solution and also why my reasoning in wrong
 
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I would try (4C3)(4C2)+(4C2)(4C3). You can just do it once and multiply by 2. Does that give the right answer?
 


no that gives the wrong answer. i need a solution with explanation not just a formula.
 


Hi batballbat! :smile:

The number of answers given is either 8, 7, 6, or 5.
How many ways to give say 6 answers?
And how many of those satisfy the condition of at least 2 in each part?
 


i didnt understand your question
 


The problem requires a candidate to give at least 5 answers.
So he could for instance give 6 answers.
At this stage I'm not looking yet whether he chooses questions from part A or from part B.
Just the total number of combinations of questions he could choose.
So he could for instance choose 1,2,3,4,5,6.
Or he could choose 1,2,5,6,7,8.
How many combinations if he chooses exactly 6 questions?
 
hi batballbat! :smile:
batballbat said:
I tried with (4 2) * (4 2) * (4 1) where (n r) is the no. of combinations of r objects from n objects. But it is the wrong answer. Give me … why my reasoning in wrong

because your (4,1) doesn't come from anywhere …

what is the group of 4 things from which 1 has to be chosen?? :confused:​
 


8c6..........
 


i know i am wrong with that reasoning. but i got (4 1) because the remaining questions is 4 after we have chosen 2 from each. And for the fifth question we have to choose one from the remaining
yes there is repetition so this does not give the answer
 
  • #10


batballbat said:
8c6..........

Yep! :smile:
(without the dots :wink:)

Now, say questions 1,2,3,4 are in part A, and questions 5,6,7,8 are in part B.
If we choose questions 1,2,3,4,5,6, then you will have 4 questions from part A and 2 questions from part B.
Which combination of questions wouldn't have at least 2 questions in each part?
 
  • #11


4.......... without the dots
 
  • #12


batballbat said:
4.......... without the dots
:rolleyes:

Can you name a specific combination of questions?
 
  • #13
batballbat said:
i know i am wrong with that reasoning. but i got (4 1) because the remaining questions is 4 after we have chosen 2 from each.

ah! i see the logic now! :smile:

nooo that doesn't work, because the new 4 isn't independent of the previous 2 …

that ()()() formula only works for separate sets

sometimes there's isn't a one-step answer to these probability questions, and you have to consider different cases

try splitting it up into two cases, the way ArcanaNoir :smile: suggested …

what do you get? :wink:​
 
  • #14


sorry my brain was repairing when i answered that
 
  • #15


i still don't have the solution
 
  • #16


Can you post the answer? Sometimes it's easier to help someone when you are sure you have it right before suggesting something. Plus, I didn't understand the question to mean you could attempt six questions. "required to attempt five" is not the same to me as "required to attempt at least five".
 
  • #17


the 4 combinations are 5678(1,2,3,4)
 
  • #18


@ARCANANOIR :smile:: Well, your answer is the right one if the candidate answers exactly 5 questions...

But as yet, I'm at a loss how to explain that.
TBH my current tactic is to let the OP think about it for a while, since he has to be able to solve this and other problems himself.

Do you want to give it a try?
 
  • #19


No, I'm satisfied knowing I got the right answer according to my interpretation of the question. I still have my own probability nightmares to work on.
 
  • #20


ArcanaNoir said:
No, I'm satisfied knowing I got the right answer according to my interpretation of the question. I still have my own probability nightmares to work on.

Hah! So you got one answer for free! :wink:


batballbat said:
the 4 combinations are 5678(1,2,3,4)

Errr... those are not combinations?

I meant combinations like:
1,2,3,4, 5,6
1,2,3,4, 5,7
3,4, 5,6,7,8
1,4, 5,6,7,8

These are 4 combinations of possible answers.
Do they satisfy the criterion at least 2 in each part?
 
  • #21


I like Serena said:
Hah! So you got one answer for free! :wink:

OP already said my answer was wrong, but maybe misinterpreted what I meant about multiplying by 2.
 
  • #22


ArcanaNoir said:
OP already said my answer was wrong, but maybe misinterpreted what I meant about multiplying by 2.

@batballbat: How did you interpret ArcanaNoir's answer?
 
  • #23


i am sorry. he was right.
 
  • #24


Well that's a relief. I thought I was doomed on my combinatorics quiz Thursday. And how sad especially after ILS taught me all that shuffling stuff this weekend!
 
  • #25


so all my helpers. i am almost going crazy with this combination problems. maybe i am missing the simple reasonings behind these problems. And i found out that this forum is not for posting solutions but for a dialectical method of finding answers by questioning. Anyways i am studying combinations from my course book, which is incomplete as me going crazy shows. so suggest me a suitable book to master this subject
 
  • #26


i am feeling i won't get an answer because dialectic method is not applicable to this question:smile:
 
  • #27


Sorry, I have no suggestions.

Back to this problem: if you have a total of 5 questions, with at least 2 in each part, that means that you have:
1) either 2 questions in part A and 3 questions in part B,
2) or 3 questions in part A nad 2 questions in part B.

So for 1) you "choose" 2 from the 4 questions in part A, and you "choose" 3 from the 4 questions in part B.
 
  • #28


batballbat said:
i am feeling i won't get an answer because dialectic method is not applicable to this question:smile:

Ah, well, I think it is applicable. :smile:

But you have to admit you threw me off by claiming that ArcanaNoir's answer was wrong, when it wasn't!

And the wording of the question is also not clear.
It doesn't say clearly whether we're talking about exactly 5 questions, or at least 5 questions.
 
  • #29


batballbat said:
i am feeling i won't get an answer because dialectic method is not applicable to this question:smile:

Maybe you should ask in the math and science learning materials section.
 

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