Calculate the number of ways of choosing $4$ numbers from the set ${1,2,\cdots ,11}$ such that at least $2$ of the numbers are consecutive.
1. **Total number of ways to choose 4 numbers from 11:**
   The total number of ways to choose 4 numbers from a set of 11 is given by the binomial coefficient:
   \[
   \binom{11}{4} = \frac{11!}{4!(11-4)!} = \frac{11!}{4! \cdot 7!} = 330
   \]

2. **Counting the cases where no numbers are consecutive:**
   To count the number of ways to choose 4 numbers such that no two are consecutive, we use a combinatorial argument. 

   - Consider the 11 numbers as 11 positions.
   - We need to choose 4 positions such that no two chosen positions are consecutive.
   - To ensure no two chosen positions are consecutive, we can place a "gap" of at least one position between any two chosen positions.

   We can think of this as choosing 4 positions out of 8 available slots (since each chosen number must be separated by at least one unchosen number). This is equivalent to choosing 4 positions from 8:
   \[
   \binom{8}{4} = \frac{8!}{4!(8-4)!} = \frac{8!}{4! \cdot 4!} = 70
   \]

3. **Complementary counting:**
   We now use complementary counting to find the number of ways to choose 4 numbers such that at least 2 of them are consecutive. This is done by subtracting the number of ways where no numbers are consecutive from the total number of ways:
   \[
   330 - 70 = 260
   \]

Conclusion:
\[
\boxed{260}
\]