91Ó°ÊÓ

We have to form $5$digit numbers using the integers $1,2,...,9$.

No digit can appear more than twice.

So, there will be three mutually exclusive even in which this can happen.

The no. of ways of choosing first digit is $\left(\begin{array}{c}9\\ 1\end{array}\right)=9$

The no. of ways of choosing first digit is $\left(\begin{array}{c}8\\ 1\end{array}\right)=8$

The no. of ways of choosing first digit is $\left(\begin{array}{c}7\\ 1\end{array}\right)=7$

The no. of ways of choosing first digit is $\left(\begin{array}{c}6\\ 1\end{array}\right)=6$

The no. of ways of choosing first digit is $\left(\begin{array}{c}5\\ 1\end{array}\right)=5$

Therefore, the number of ways in which all $5$ digits appear only once =$9×8×7×6×5=15120$

No. of ways in which we can select the $1$digit that will appear twice is $=\left(\begin{array}{c}9\\ 1\end{array}\right)=9$

No. of ways in which we can select 2 spots to place the 1 digit that will appear twice is $=\left(\begin{array}{c}5\\ 2\end{array}\right)=10$

The no. of ways of selecting $3$digits out of the remaining $8$digits is $=8×7×6=336$

Therefore, the number of ways in which one digit appears twice $=9×10×336=30240$

No. of ways in which we can select the $2$digits that will appear twice is$=\left(\begin{array}{c}9\\ 2\end{array}\right)=36$

No. of ways in which we can select $4$spots to place the $2$digits that will appear twice is$=\left(\begin{array}{c}5\\ 4\end{array}\right)=5$

The no. of ways to place the selected $2$digits at $4$spots such that doubles can be avoided is$=\frac{4!}{2!2!}=6$

The no. of ways of selecting 5th digit out of the remaining 7 digits is $=\left(\begin{array}{c}7\\ 1\end{array}\right)=7$

Therefore, the number of ways in which two digits appear twice$=36×5×6×7=7560$

Therefore, the no. of $5$-digits numbers that can be formed are$=15120+30240+7560=52920$