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Consider $30$ multiple-choice questions, each with four options of which exactly one is correct. Then the number of ways one can get only the alternate questions correctly answered is

  1. $3^{15}$
  2. $2^{31}$
  3. $2 \times \begin{pmatrix} 30 \\ 15 \end{pmatrix}$
  4. $2 \times 3^{15}$
in Combinatory by Veteran (424k points)
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@techbd123

Hey would you like to try this que?

+1

The question seems a bit ambiguous to me by the term "only the alternate questions correctly". What does it mean actually? Does it mean like this below?


CA= Correctly Answered
IA= Incorrectly Answered
NA=Not Answered

1 2 3 4 5 6 ... 28 29 30
CA IA or NA CA IA or NA CA IA or NA ... IA or NA CA IA or NA

Here's $2^{15}$ cases.
Again

1 2 3 4 5 6 ... 28 29 30
IA or NA CA IA or NA CA IA or NA CA ... CA IA or NA CA

Here's $2^{15}$ cases.


Total $=2^{15}+2^{15}=2(2^{15})=2^{16}$

+1
Yeah, same doubt I had as well.
+1
Two ways for answering all the questions such that alternate answers are correct--

$T F T F \ldots TF$

$FTFT\ldots FT$

==> $3^{15}+3^{15}$
+1

Why $3$ as the base?

@Verma Ashish

0
There are 3 ways to choose wrong option, for each question.

And only 1 way to choose correct option Because only one choice is correct out of four options.
+1
Then how would you explain this statement?

Then the number of ways one can get only the alternate questions correctly answered is

The question asks for true answers, right?

You considered for false only?
+1
There is only one correct option for right answer, whereas there were 3 different ways to answer it wrong. Since you had to get alternate questions right, you could start with either a right answer or wrong. Can you please elaborate your doubt?
0

@imShreyas

I thought in a same way but i have  doubt..

This is the way i solved the problem:

let 1,2,3,4...30 be the questions and each question has options a,b,c,d.

to satisfy the question student must answer,in one of following ways.

1,3,5,7,9,...29

or

2,4,6,8,...,30

why only these 2 sequences?

Let us start with 3 and answering alternate questions correctly.. then we have sequence 3,5,7,...29 but we have solved only 14 questions so he has to choose one number from  1 or 2 but only 1 satisfy the condition.

So the sequence formed is again 1,3,5,...29

Let us start with 9 and answering alternate questions correctly.. then we have sequence 9,11,13,15...29 but we have solved only 11 questions so he has to choose 4 numbers from 1 to 8  but only 1,3,5,7 satisfy the condition.

So the sequence formed is again 1,3,5,...29

similarly we can prove for 2,4,6,...30.

So even if we try to form a different sequence, we end up only in one of these 2 sequences..

and Now consider the sequence 1,3,5,....29

each of this question has to be answered correctly and all other questions has to be wrong.

so each of 1,3,5...29 is to be answered in 1 way and each of 2,4,6...30 has 3 choices to be answered..

so we have $3^{15}$ ways.

similarly we have $3^{15}$ ways for other sequence..

So on total there are $3^{15}+3^{15}$ ways.. which is option c

but my doubt is why are we not considering the order in which questions are answered.

I mean each sequence can be answered in $15!$ ways..

so why can't the answer be $2*3^{15}*15!$

where am i going wrong?

 

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