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Let's first look at 3$3$ digit phone numbers to simplify. If the first digit is "locked" at 0$0$ only, we then have 10$10$ choices for the 2$2$nd digit and 10$10$ choices for the 3$3$rd digit, thus giving us 10$10$ x 10$10$ = 102${10}^2$ = 100$100$ possible phone numbers. Examples would be 000,001,012...010,011,012...090,091,092...099$000,001,\mathrm{012...}010,011,\mathrm{012...090},091,\mathrm{092...099}$

For 10$10$ digit phone numbers with the first digit "locked" at 0$0$ it is similar except it would be 109${10}^9$ = 1,000,000,000$1,000,000,000$ = 1$1$ billion.

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edited Nov 17 '14 at 10:44

answered Nov 17 '14 at 10:39

David

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https://math.stackexchange.com/questions/1025767/number-of-different-possible-permutations-of-a-telephone-number

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