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- First digit . . . If it can't be 0, 1, or 8, then there are 7 choices.

- Second digit . . . Since no repeats are permitted, there are 9 choices ... 10 digits but not the same as the first one.

- Third digit . . . any digit that's not the same as either of the others . . . 8 choices.

Which means there are (7 x 9 x 8) = 504 different possible area codes.

HOWEVER . . . as to how much freedom a person has to "choose" one . . . That decision is typically the choice of your telephone service provider, often but not always corresponding to the geographic location to which his phone number is registered.

So the strict answer to the question may well be "One, unless the person is willing to move."

(Also, note that the restrictions in this homework problem are not the same as the restrictions on real area codes. There are 648 possible area codes in the North American system, not counting special codes like toll-free 800/888/etc.)

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10y ago
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10y ago

As your question is written, there are 720 possibilities, 10 x 9 x 8. There are 10 possibilities for the first digit. The second digit cannot be the same as the first, so that leaves 9 possibilities. The third digit cannot be the same as the first or second, so that leaves 8 possibilities.

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Q: How many area codes would be possible if all three digits could be any value 1-8?
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