# In how many ways one can answer a 20-question exam with two-choice questions?

There is a problem in my book:

In how many ways one can answer a 20-question exam with two-choice questions?

which I would solve as follows:

Since there are 20 questions, and each question has two choices, there are $20 \cdot {{2} \choose 1}$ but I am wrong according to the solution written in the book. Why I can't do it this way? I know what the correct answer is, I just want to know why my solution doesn't work in this case.

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Let's see what $20\binom{2}{1}$ actually counts. For any question, like the $7$-th, you have $\binom{2}{1}$ choices. So there are $\binom{2}{1}$ ways to answer the $7$-th question and leave all the others blank. Now your $20\binom{2}{1}$ counts all the different ways to answer exactly one of the $20$ questions and leave the others blank.

My candidate for the right answer is $3^{20}$. For at any question, we can choose the first answer, or the second, or decide the question is too hard and go on to the next question.

But the book probably gave $2^{20}$. Think of choosing the first answer to a question as the letter $a$, and choosing the second answer as the letter $b$. Then the number of ways to answer every question is just the number of "words" of length $20$ over the alphabet $\{a,b\}$.

One useful way to see whether one has the right answer is to do a hand count in "small" cases, and check one's conjectured formula against that count. For $n=1$, your method gives $2$, which is correct. For $n=2$, your method gives $4$, which is correct. Good so far. For $n=3$, your method gives $6$, which is not right: listing gives $8$ ways.

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Think of a three question test. For each answer on question 1 you have two choices for question 2, for a total of 4. Now for each of the 4 you have 2 choices for question 3, making 8. So the choices multiply.

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