Question

How many different seven-digit numbers can be written using only three digits, 1, 2, and 3, under the condition that the digit 2 occurs twice in each number?

Answer 1

The different scenario with frequency of $1,2,3$ in a $7$ digit numbers are shown below$.$

Number$:$ $1,2,3$

Frequency$:$ $053,143,323,233,41,3,503.$

$\therefore$ The no. of different $7$ digits is

$=\frac{7!}{0!5!3!}$$+\frac{7!}{1!4!3!}$$+\frac{7!}{2!3!3!}$$+\frac{7!}{2!3!3!}$$+\frac{7!}{4!1!3!}+\frac{7!}{5!0!2!}$$=7+35+70+70+35+21=238$