Given that, the digits are $1,2, \text{and} \; 3.$
There are different possibilities.
- One digit is repeated, and the other two digits are not repeated.
- $1\quad1\quad2\quad3$
- $1\quad2\quad2\quad3$
- $\underbrace{1\quad2\quad3\quad3}_{\dfrac{4!}{2!} \times 3 = 36}$
- Two digits repeated.
- $\underbrace{2\quad2\quad3\quad3}_{\dfrac{4!}{2!2!} = 6}$
- One digit repeated three times.
- $2\quad3\quad3\quad3$
- $\underbrace{2\quad2\quad2\quad3}_{\dfrac{4!}{3!} \times 2 = 8}$
$\therefore$ The total number of four-digit numbers $= 36+6+8 = 50$
Correct Answer $:50$