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A code consists of 5 digit numbers arranged from integers $0, 1, 2, ..., 9$. How many code with no two digits side by side are same?

I have tried to answer it as below.

First digit can filled by $0, 1, ...,9$ so it have $10$ options.

Second digit can filled by $0, 1, ...,9$ except the digit used in first digit, so it have $9$ options.

Third digit can filled by $0, 1, ...,9$ except the digit used in second digit, so it have $9$ options.

Fourth digit can filled by $0, 1, ...,9$ except the digit used in third digit, so it have $9$ options.

Fifth digit can filled by $0, 1, ...,9$ except the digit used in fourth digit, so it have $9$ options.

So, the number of code is $$10\times 9^4.$$ I'm not sure with my answer. Is it right answer? If my answer is wrong, what the hint to be used for answer this problem?

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You are correct.

If you have doubt, a generally good idea (esp. in combinatorics) is to try smaller examples. E.g. length-$3$ codes using just letters A,B,C. By your logic you should have $3 \times 2 \times 2 = 12$ such codes (out of $27$ possible). This is easy to check by hand:

AAA, AAB, AAC, ABA, ABB, ABC, ACA, ACB, ACC,

BAA, BAB, BAC, BBA, BBB, BBC, BCA, BCB, BCC,

CAA, CAB, CAC, CBA, CBB, CBC, CCA, CCB, CCC.

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