Answer:
There are 49,000,00,000 possible numbers.
49,000,00,000 números diferentes posibles.
Explanation:
If we have n trials, each with m possible outcomes, the total number of possible outcomes is:
![T = m^(n)](https://img.qammunity.org/2021/formulas/mathematics/college/uds54vf4y16ttcvxwy0sawkbarpbui671c.png)
In this question:
The first two trials(numbers), 3 digits are not possible in each. So 7 are possible.
The last eight numbers, all 10 digits are possible.
Then
![T = 7^(2)*10^(8) = 49,000,00,000](https://img.qammunity.org/2021/formulas/mathematics/high-school/sdip5soyi1z7ti5poc938vqfs9kporrcf7.png)
There are 49,000,00,000 possible numbers.
49,000,00,000 números diferentes posibles.