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A student is taking a $12$-question multiple choice test. Each question has $5$ choices. How many ways can the student fill out the answer sheet so there is at least one place where two consecutive answers are the same?

This is what I did:

Total possible ways the student can fill the answer sheet: $5^{12}$ (because he has $5$ options in the $12$ questions).

Total possible ways the student can fill the answer sheet with no consecutive answers: $5 \cdot 4^{11}$ (In the first question he can answer any of the $5$ options, however, in the remaining questions he can only answer $4$ options to avoid choosing consecutive answers.)

Total possible ways the student can fill the answer sheet with at least two consecutive answers: $5^{12}- 5 \cdot 4^{11} = 223,169,105$.

I doubt about the total possible ways the student can fill the answer sheet with no consecutive answers... is my logic right?

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  • $\begingroup$ Write out a detailed explanation of the $5\times4^{11}$. This should either convince you that you are right, or show you where you are wrong. Do this for the first part of your answer too. $\endgroup$ – David May 7 '19 at 5:17
  • $\begingroup$ Sorry, I had a typo on my first statement. $\endgroup$ – Nico May 7 '19 at 5:33
  • $\begingroup$ Yes, you are correct. $\endgroup$ – N. F. Taussig May 7 '19 at 9:05
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You are correct, $5\times 4^{11}$ is the number of ways to fill out the test with no two consecutive equal answers, so $5^{12}-5\times 4^{11}$ is the number of ways to fill out the test so that there are two consecutive equal answers.

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I am trying to do this question and am getting a different answer. Instead of making a new thread, I hope I am allowed to post my response and see where people think I'm going wrong here. If not please let me know and I will delete.

Denote the answers of the multi-choice test to be A_1,A_2,..., A_12.

Now, for us to have consecutive answers be the same it must be that A_1 = A_2 or A_2 = A_3 or .... or A_11 = A_12. Denote this pair of questions with the same answer to be A_s.

Now, consider a new set which contains 11 elements: A_s, and our other 10 remaining answers to questions. Certainly, we can position A_s anywhere, as remember that we have set A_s to be the two consecutive answers already. Thus, we have no conditions on the remaining choices. We can re-arrange this set 11! different ways (as it is simply the re-arrangements of a set with 11 elements where order matters), and each element can have 5 different options.

Thus, we have 11!x(5 x 11) = 21,954,424,000 different options.

My answer seems way too big

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  • $\begingroup$ You have not accounted for the possibility that there is more than one pair of consecutive answers that are the same. To do so, you could apply the Inclusion-Exclusion Principle, but with $11$ possible pairs of consecutive answers that could be the same, that could prove exceedingly difficult. $\endgroup$ – N. F. Taussig May 20 '19 at 0:02

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