The number $2013$ has the property that it includes four consecutive digits ($0$, $1$, $2$, and $3$). How many $4$-digit numbers include $4$ consecutive digits?
[i](9 and 0 are not considered consecutive digits.)[/i]

$\text{(A) }18\qquad\text{(B) }24\qquad\text{(C) }144\qquad\text{(D) }162\qquad\text{(E) }168$
1. **Identify the possible sets of four consecutive digits:**
   The sets of four consecutive digits are:
   \[
   (0, 1, 2, 3), (1, 2, 3, 4), (2, 3, 4, 5), (3, 4, 5, 6), (4, 5, 6, 7), (5, 6, 7, 8), (6, 7, 8, 9)
   \]
   This gives us 7 possible sets.

2. **Calculate the number of permutations for sets without 0:**
   For the 6 sets that do not include 0, each set can be arranged in \(4!\) ways:
   \[
   4! = 24
   \]
   Therefore, for these 6 sets, the total number of permutations is:
   \[
   6 \times 24 = 144
   \]

3. **Calculate the number of permutations for the set with 0:**
   For the set \((0, 1, 2, 3)\), 0 cannot be the thousands digit. Therefore, we need to arrange the digits such that 0 is not the first digit. The number of valid permutations is:
   \[
   3! = 6
   \]

4. **Sum the total number of valid 4-digit numbers:**
   Adding the permutations from the sets without 0 and the set with 0, we get:
   \[
   144 + 6 = 150
   \]

5. **Verify the calculation:**
   The solution provided in the community suggests:
   \[
   7(4!) - 3! = 7 \times 24 - 6 = 168 - 6 = 162
   \]
   This calculation is incorrect because it does not properly account for the restriction on the set with 0.

The final answer is \(\boxed{150}\)