Mary typed a six-digit number, but the two $1$s she typed didn't show.  What appeared was $2002$.  How many different six-digit numbers could she have typed?

$\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }15\qquad\textbf{(E) }20$
To solve this problem, we need to determine the number of different six-digit numbers Mary could have typed, given that the two $1$s she typed didn't show up in the displayed number $2002$.

1. **Identify the positions for the missing $1$s:**
   The displayed number is $2002$, which has four digits. We need to place two $1$s into this number to form a six-digit number. This means we need to find positions for the two $1$s among the six possible positions in the six-digit number.

2. **Calculate the number of ways to place the $1$s:**
   We can place the two $1$s in two different ways:
   - Both $1$s are placed in different positions.
   - Both $1$s are placed in the same position.

3. **Case 1: Placing the $1$s in different positions:**
   We need to choose 2 out of the 6 positions to place the $1$s. This can be done in $\binom{6}{2}$ ways.
   \[
   \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15
   \]

4. **Case 2: Placing the $1$s in the same position:**
   This case is not possible because placing both $1$s in the same position would not increase the number of digits in the displayed number. Therefore, this case does not contribute to the count.

5. **Combine the results:**
   Since placing both $1$s in the same position is not possible, the total number of different six-digit numbers Mary could have typed is given by the number of ways to place the $1$s in different positions, which is $15$.

\[
\boxed{15}
\]