Number of ways an answers sheet can be fill with at least two consecutive answers? 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? 
 A: 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. 
A: 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 
