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For years, telephone area codes in the United States and Canada consisted of a sequence of three digits. The first digit was an integer between 2 and 9, the second digit was either 0 or 1, and the third digit was any integer from 1 to 9. How many area codes starting with a 4 were possible?

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musiclover10045

The first number had to be 4, so only one option for the first number.

The second number was either 0 or 1, so two options for the second number.

The third number was between 1 and 9, so nine options for the third number.

Multiply each number of options together for total options:

1 x 2 x 9 = 18 different codes could start with 4.

MrRoyal

The question illustrates permutation and combination.

The total number of area codes starting with 4 is 18.

The selection is as follows:

  • The first digit of the code must start with 4 (so, the number of selection is 1)
  • The second digit is either 0 or 1 (so, the number of selection is 2)
  • The third digit is any of 1 - 9 (so, the number of selection is 9)

The total number of selection is:

[tex]n = 1 \times 2 \times 9[/tex]

[tex]n = 18[/tex]

Hence, 18 area codes starting with 4 are possible.

Read more about permutation and combination at:

https://brainly.com/question/15301090

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