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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?

User Krisna
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1 Answer

4 votes

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)

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.

User Xgo
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