jmc

Suppose that the two identical digits are both $1$. Since the thousands digit must be $1$, only one of the other three digits can be $1$. This means the possible forms for the number are $11xy,1x1y,1xy1$ Because the number must have exactly two identical digits, $x\ne y$, $x\ne 1$, and $y\ne 1$. Hence, there are $3\cdot 9\cdot 8=216$ numbers of this form. Now suppose that the two identical digits are not $1$. Reasoning similarly to before, we have the following possibilities: $1xxy,1xyx,1yxx.$ Again, $x\ne y$, $x\ne 1$, and $y\ne 1$. There are $3\cdot 9\cdot 8=216$ numbers of this form. Thus the answer is $216+216=\boxed{432}$.