An ant climbs either two inches or three inches each day. In how many ways can the ant climb twelve inches, if the order of its climbing sequence matters?
    
$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }14$
To determine the number of ways the ant can climb twelve inches, we need to consider the different combinations of 2-inch and 3-inch climbs that sum to 12 inches. We will analyze the problem by considering the number of days the ant climbs and the combinations of 2-inch and 3-inch climbs.

1. **Case 1: The ant climbs 12 inches in 4 days.**
   - The only way to achieve this is by climbing 3 inches each day.
   - Sequence: $(3, 3, 3, 3)$
   - Number of ways: $1$

2. **Case 2: The ant climbs 12 inches in 5 days.**
   - The ant must climb 3 inches on exactly 2 of the days and 2 inches on the remaining 3 days.
   - We need to choose 2 days out of 5 for the 3-inch climbs.
   - Number of ways: $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = 10$

3. **Case 3: The ant climbs 12 inches in 6 days.**
   - The only way to achieve this is by climbing 2 inches each day.
   - Sequence: $(2, 2, 2, 2, 2, 2)$
   - Number of ways: $1$

Adding up all the possible ways from each case, we get:
\[ 1 + 10 + 1 = 12 \]

Thus, the total number of ways the ant can climb twelve inches is $\boxed{12}$.