# How many code with no two digits side by side are same?

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?

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: