Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

(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


For part (a) I've got the answer, it is $2^{10}$.

For part (b) I think the answer is $ 10 \times 2^9 $ because the number of ways to choose the question to answer is 10 and in each selection the number of ways to answer the question is $2^9$ but the answer provided in the book is $3^{10}$.

Can someone explain to me?

share|improve this question
2  
Probably the book suggest that the student can leave as many questions unanswered as he wants. –  barto Feb 10 '13 at 14:43
    
@barto. I agree. On (b) she/he has 3 possibilities to each question. So $3^{10}$. –  Sigur Feb 10 '13 at 14:44
    
For your response to part b you should have $10 \times 2^9 + 2^{10}$ since "it is possible to leave a question unanswered" does not require an unanswer. But it is more likely to mean there can be any number of unanswers from 0 through to 10. –  Henry Feb 10 '13 at 15:08

2 Answers 2

If the answer has to be $3^{10}$, then this means that in case (b) it is intended that for each question the student can choose 1`out of 3 possibilites: true, false, not telling.

share|improve this answer

In part b), for simplicity, 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 $2^1$ ways in which they can answer question 2, and vice versa. Thus there are $2\times 2^1$ 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 $2^2 + 2\times 2^1 + 1 = (2+1)^2 = 3^2$ ways of answering the questions.

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.