We see that the possible such arrangements fall into cases:
- We can replace the $C$ by a $M$ or $W$ and the arrangement is still valid
- We can't replace the $C$ by a $M$ or $W$ and keep the arrangement valid.
In the first case, once we do the replacement, we have only one valid arrangement: $MWM\cdots WM$ (dually, $WMW\cdots MW$) and so we see that there are 11 such arrangements once we remember that some $M$ (or $W$) is actually C, so 11 in each subcase for a total of 22.
In the second case, each such arrangement is $MWMW\cdots MW$ or $WMWM\cdots WM$ with a $C$ added somewhere that isn't one of the ends. This gives us 19 possible places to add a $C$ in each subcase, or 38 total arrangements in this case.
So there are 60 ways to arrange the $M$s, $W$s, and $C$.
If the adults are distinguishable, we've forgotten that order matters. So we need to multiply by all the ways to arrange the 10 $M$s and 10 $W$s. But that's just $10!$ for each, so our final answer is $60(10!)^2$.