The familiar $3$-dimensional cube has $6$ $2$-dimensional faces, $12$ $1$-dimensional edges, and $8$ $0$-dimensional vertices. Find the number of $9$-dimensional sub-subfaces in a $12$-dimensional cube.
1. **Understanding the problem**: We need to find the number of 9-dimensional sub-subfaces in a 12-dimensional cube. A sub-subface of a cube is a lower-dimensional face of the cube. For example, in a 3-dimensional cube (a regular cube), a 2-dimensional subface is a square face, a 1-dimensional subface is an edge, and a 0-dimensional subface is a vertex.

2. **Generalizing the concept**: In an \(n\)-dimensional cube, a \(k\)-dimensional subface is formed by fixing \(n-k\) coordinates to either 0 or 1 and allowing the remaining \(k\) coordinates to vary. 

3. **Choosing the dimensions**: To form a 9-dimensional subface in a 12-dimensional cube, we need to choose 9 dimensions out of the 12 to vary. The number of ways to choose 9 dimensions out of 12 is given by the binomial coefficient:
   \[
   \binom{12}{9}
   \]
   Using the property of binomial coefficients, we know that:
   \[
   \binom{12}{9} = \binom{12}{3}
   \]
   This is because \(\binom{n}{k} = \binom{n}{n-k}\).

4. **Calculating the binomial coefficient**:
   \[
   \binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12!}{3! \cdot 9!}
   \]
   Simplifying the factorials:
   \[
   \binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220
   \]

5. **Counting the subfaces**: Each choice of 9 dimensions corresponds to a 9-dimensional subface. For each of these choices, the remaining 3 dimensions can each be fixed to either 0 or 1, giving \(2^3 = 8\) possible ways to fix the remaining dimensions.

6. **Multiplying the results**: Therefore, the total number of 9-dimensional subfaces is:
   \[
   8 \times 220 = 1760
   \]

The final answer is \(\boxed{1760}\).