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QUESTION 12
How many nine-digit ZIP code numbers are possible if the first digit cannot be a three and successive digits cannot be the same?
10^9 =1,000,000,000
9^9 =387,420,489
362,880
3,628,800

Respuesta :

Using the Fundamental Counting Theorem, it is found that the number of possible ZIP codes is:

[tex]9^9 =387,420,489[/tex]

Fundamental counting theorem:

States that if there are n things, each with [tex]n_1, n_2, …, n_n[/tex] ways to be done, each thing independent of the other, the number of ways they can be done is:

[tex]N = n_1 \times n_2 \times \cdots \times n_n[/tex]

In this problem:

  • The first digit cannot be 3, hence there are 9 possible outcomes for the first digit, that is, [tex]n_1 = 9[/tex].
  • For the next 8 digits, there are also 9 possible outcomes, as it cannot repeated the previous digit, hence [tex]n_2 = n_3 \cdots n_9 = 9[/tex]

Hence:

[tex]N = n_1 \times n_2 \times \cdots \times n_9 = 9^9 = 387,420,489[/tex]

A similar problem is given at https://brainly.com/question/19022577