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Combinatorics question

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Combinatorics question [#permalink]

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New post 30 Aug 2017, 00:39
Can someone explain the following solution for this question? If a student doesn't answer the question, then there should be 2^1 ways, but why is it regarded as 2*2^1 ways?

---
In how many ways can the students answer a 10-question true false examination?

(a) In how many ways can the students answer a 10-question true false examination?
(b). In how many ways can the student answer the test in part (a) if it is possible to leave a question unanswered in order to avoid an extra penalty for a wrong answer

---
The explanation is:
"let's reduce the problem to just two questions.

There are 2^2 ways in which both questions may be answered true/false.

If a student does not answer question 1, there are still 2*2^1 ways in which they can answer question 2, and vice versa. Thus there are 2*2^1 ways in which only one question is answered.

Finally, there is just one way in which neither question is answered.

Putting this together, there are 2^2+2*2^1+1=(2+1)^2=3^2=32 ways of answering the questions.

Extending this to ten questions, there are 2^10 ways to answer all ten questions, 10*2^9 for answering all but one question, 45×2^8 ways of answering all but two questions, etc, giving a total of (2+1)^10=3^10

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Re: Combinatorics question [#permalink]

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New post 01 Sep 2017, 10:57
annguyenup wrote:
Can someone explain the following solution for this question? If a student doesn't answer the question, then there should be 2^1 ways, but why is it regarded as 2*2^1 ways?

---
In how many ways can the students answer a 10-question true false examination?

(a) In how many ways can the students answer a 10-question true false examination?
(b). In how many ways can the student answer the test in part (a) if it is possible to leave a question unanswered in order to avoid an extra penalty for a wrong answer

---
The explanation is:
"let's reduce the problem to just two questions.

There are 2^2 ways in which both questions may be answered true/false.

If a student does not answer question 1, there are still 2*2^1 ways in which they can answer question 2, and vice versa. Thus there are 2*2^1 ways in which only one question is answered.

Finally, there is just one way in which neither question is answered.

Putting this together, there are 2^2+2*2^1+1=(2+1)^2=3^2=32 ways of answering the questions.

Extending this to ten questions, there are 2^10 ways to answer all ten questions, 10*2^9 for answering all but one question, 45×2^8 ways of answering all but two questions, etc, giving a total of (2+1)^10=3^10



Quote:
(a) In how many ways can the students answer a 10-question true false examination?


Let's assume there are 3 questions. The answer can be true or false. you can use slot method:

2 2 2 - so there are 2 outcomes for the first question, 2 outcomes for the second, 2 outcomes for the third:2x2x2=\(2^3\)

8 outcomes overall :

true true true
true true false
true false true
true false false
false false true
false true false
false true true
false false false

Quote:
(b). In how many ways can the student answer the test in part (a) if it is possible to leave a question unanswered in order to avoid an extra penalty for a wrong answer


same idea here, but there are three outcomes possible. If there are 3 questions, then 3 x 3 x 3 =3^3
If there are ten questions, then \(3^10\)
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Re: Combinatorics question [#permalink]

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New post 06 Sep 2017, 03:48
I don't quite understand your question - do you still need some explanation (solution) of the problem to the already existing one? If you're in doubt about solution, you can get more help on the website, where experts are working in different areas and in particular in the field of combinatorics.

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Combinatorics question [#permalink]

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New post 07 Sep 2017, 12:47
annguyenup wrote:
Can someone explain the following solution for this question? If a student doesn't answer the question, then there should be 2^1 ways, but why is it regarded as 2*2^1 ways?

---
In how many ways can the students answer a 10-question true false examination?

(a) In how many ways can the students answer a 10-question true false examination?
(b). In how many ways can the student answer the test in part (a) if it is possible to leave a question unanswered in order to avoid an extra penalty for a wrong answer

---
The explanation is:
"let's reduce the problem to just two questions.

There are 2^2 ways in which both questions may be answered true/false.

If a student does not answer question 1, there are still 2*2^1 ways in which they can answer question 2, and vice versa. Thus there are 2*2^1 ways in which only one question is answered.

Finally, there is just one way in which neither question is answered.

Putting this together, there are 2^2+2*2^1+1=(2+1)^2=3^2=32 ways of answering the questions.

Extending this to ten questions, there are 2^10 ways to answer all ten questions, 10*2^9 for answering all but one question, 45×2^8 ways of answering all but two questions, etc, giving a total of (2+1)^10=3^10


The language in the explanation is slightly ambiguous. By saying "2*2^1 ways in which only one question is answered," they aren't referring to the number of ways to answer any particular single question. (You're correct, by the way - there are two ways to answer a single question, assuming you don't count 'no answer' as a type of answer.)

What they appear to mean is actually that there are four different scenarios (2*2^1) for a two-problem test, in which exactly one of the two problems is answered and the other isn't. Here are those four scenarios:

question 1, question 2
no answer, true
no answer, false
true, no answer
false, no answer
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Combinatorics question   [#permalink] 07 Sep 2017, 12:47
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