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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 $:50$