The possible ways to create an acceptable string of length $k$ are:
- Start with an acceptable string of length $k-3$, and take the $D\to ACD$ path.
- Start with an acceptable string of length $k-3$, and take the $D\to ABD$ path.
- Start with an acceptable string of length $k-2$, and take the $D\to BD$ path.
Thus, we achieve the recurrence relation given below:
$$N(k) = \underbrace{N \Bigl (k-3 \Bigr )}_{D\to ACD} + \overbrace{N \Bigl (k-3 \Bigr )}^{D\to ABD} + \underbrace{N \Bigl (k-2 \Bigr )}_{D\to BD}$$
$$\boxed{\displaystyle N(k) = 2N \Bigl (k-3 \Bigr ) + N \Bigl (k-2 \Bigr )}$$
The sequence for $N(k)$ thus generated is
$$\small 0, 0, 2, 0, 2, 4, 2, 8, 10, 12, 26, 32, 50, 84, 114, 184, 282, 412, 650, 976, 1474, \ldots$$
http://ideone.com/JtmMW1
So, $N(11)=32$ is the correct answer.