Here's my attempt:

There are 2 cases to consider: exactly 2 C's are next to each other, or 3 C's are next to each other.

If 3 C's are next to each other, there are 6 ways to put them. Then, for the remaining 5 digits, there are \({5! \over 2!} = 60\) ways to order them. So, there are \(6 \times 60 = 360\) ways to order this.

If 2 C's are next to each other, then there are 2 subcases: The 2 C's are on the sides (1 and 2 or 7 and 8) or they are in the center.

If the C's are on the sides, there are 2 ways to choose to put the C's. Then, the third C has 5 choices (can't be right next to the other C's). Then, there are \({5! \over 2!} = 60\) ways to order the remaining digits. So, there are \(2 \times 5 \times 60 = 600\) ways for this case.

Now, if the C's are in the center, there are 5 ways to put the C's. Then, the third C has 4 choices (can't be on either side). Then, there are \({5! \over 2!} = 60\) ways for the remaining digits. So, there are \(5 \times 4 \times 60 = 1200\) ways for this.

So, there are \(3360 - 300-600-1200 = \color{brown}\boxed{1260}\) ways. I think...