Answer:
There are 49,000,00,000 possible numbers.
49,000,00,000 números diferentes posibles.
Step-by-step explanation:
If we have n trials, each with m possible outcomes, the total number of possible outcomes is:
[tex]T = m^{n}[/tex]
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
[tex]T = 7^{2}*10^{8} = 49,000,00,000[/tex]
There are 49,000,00,000 possible numbers.
49,000,00,000 números diferentes posibles.
