Combinatorics

Question

A proffesor writes 40 discrete mathematics true/false question . of the statements in these question 17 are true . if the question can be positioned in any order . then how many diff key are possible ?

Answer 1

Let us consider one instance of arrangement of question.

So in this out of 40 , 17 is going to be true and remaining false.

So it is a problem of selection(combination) since the arrangement among the questions having "true" as answer will be counted 1 only and similarly for false.

So we have to select 17 out of 40 questions , the remaining are going to have "false" as the answer.

So no of ways to do this = ⁴⁰C₁₇ which is hence the number of keys possible for one arrangement of questions.

Now this is for one arrangement of question sequence.

But given in the question we can have questions positioned in any order , so that needs to be taken into account.So ordering of questions matters .Also the questions will be distinct obviously.

Hence no of such question arrangements of 40 questions = 40!

Hence total no of keys possible overall = 40! * ⁴⁰C₁₇

Comments

@habib ,can you please explain what is meant by number of keys possible??i did not get the meaning.

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okay,we can clear it with 3 questions..where 1 is correct ,

now it can be Q1=t,Q2=f,Q3=F or Q1=F,Q2=T,Q3=F or Q1=F,Q2=F,Q3=T

so 3 ways to make 1 answer correct! now we can arrange 3 questions in any way like q1,q2,q3 or q3,q1,q2 .... in this way=3!

 

so 3!*3c1

so habib is correct

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If the question wanted to ask that for any 17 question, the statements are true ?  so 40C17

or

 

is it decided for which 17 question it would be true?