Answer:
768 numbers
Explanation:
Given
![Digits = \{0,1,2,3,4,5,6,7,8,9\}](https://img.qammunity.org/2022/formulas/mathematics/high-school/nmpvbkxf289mvj0ctlizhzj6wycknm2jk7.png)
Required
Number of digits between 0 ad 9999 that have one 1 and one 3
There are a total of 4 digits that make up any of the numbers in 0 to 9999 with the given condition
This can be represented as WXYZ
- Digit 1 can occupy any of the 4 positions
- Digit 3 can occupy any of the 4 - 1 positions
The remaining 8 digits will occupy the last 2 positions in the following ways:
- The first of the 8 digits can be selected from any of
i.e. 8 ways
- The second can be selected from any of
i.e. 8 ways
So, the total number of selection is:
![Total = 4 * (4 - 1) * 8 * 8](https://img.qammunity.org/2022/formulas/mathematics/high-school/iqlqhayti8fll8lcnidxifmkg7jjhi9s0y.png)
![Total = 4 * 3 * 8 * 8](https://img.qammunity.org/2022/formulas/mathematics/high-school/fw2x5dpqtxdmb7u6q0phu42lu3g7cmw57c.png)
![Total = 768](https://img.qammunity.org/2022/formulas/mathematics/high-school/7ndnqd32b3g24sdfno9kwl0avi1li23fpm.png)