Here is the question:
In how many ways we can construct a 11-digit long string that contains all 10 digits without 2 consecutive same digits.
Initially, I came up with $10!9$. I thought that there are $10!$ ways to construct 10-digit number with all 10 digits. And I can add one more digit at the end of each number in $9$ ways.
However, I found that may wrong. Because I when applied the same rule to 4-digit number with 3 digits(0,1,2), the answer is not $3!2$. For example, it doesn't contain $1210, 2120, 0102, ...$
So how to approach this problem?