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?