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A four-digit number is formed by using only the digits $1, 2$ and $3$ such that both $2$ and $3$ appear at least once. The number of all such four-digit numbers is
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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 $:50$

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