How many five digit numbers can be made having exactly two identical digits? 
 

$$\begin{gathered} \text{Case I:}\text{Two identical digits are }0,0\text{. } \\ {\text{ The number of ways to select three more digits is }}^9{C}_3\text{. The number of } \\ \text{ arrangements of these five digits is }5!/2!-4!=36\text{. } \\ \text{ Hence, the number of such numbers is }{ \\ }^9{C}_3\times 36=3024 \end{gathered}$$

$$\begin{gathered} \text{ Case II }\text{: Two identical digits are }(1,1)\text{ or }(2,2)\text{ or .. ог }(9,9)\text{. } \\ \mathbf{\text{ If }}\mathbf{0}\mathbf{\text{ is included,}}{\text{ then number of ways of selection of two more digits is }}^8{C}_2\text{. The } \\ \text{ number of ways of arrangements of these five digits is }5!/2!-4!2!=48\text{. } \\ \mathbf{If}\mathbf{ }\mathbf{0}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{not}\mathbf{ }\mathbf{included} \mathrm{then} \mathrm{number}\mathrm{of} \mathrm{ways} \mathrm{of} \mathrm{selection} \mathrm{of} \mathrm{three} \mathrm{more} \mathrm{digits} \mathrm{is}{ }^8{C}_3 \\ \text{. Therefore, number of such numbers is }{ \\ }^8{C}_3\times 5!/2!{=}^8{C}_3\times 60\text{. Hence, total number of five-digit numbers with } \\ \text{ identical digits }(1,1)\text{ or }(2,2)\text{ or }..(9,9)\text{ is } \\ 9\times \left({ }^8{C}_2\times 48{+}^8{C}_3\times 60\right)=42336 \\ \text{ the required number of numbers is }3024+42336=45360 \end{gathered}$$