Question

Los números de teléfonos celulares constan de 10 dígitos, ¿cuántos números diferentes se pueden tener si el primer dígito no debe ser 0, 2 ni 4 y el segundo 4, 6 ni 8?

Answer 1

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:

$T = m^{n}$

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

$T = 7^{2}*10^{8} = 49,000,00,000$

There are 49,000,00,000 possible numbers.

49,000,00,000 números diferentes posibles.