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(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?

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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

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.

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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 $(2+1)^{10} = 3^{10}$.

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