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Counting phone numbers

  1. Nov 12, 2004 #1

    THA

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    Problem details:
    I think if I pretend that N is also a number from 0-9, then there would be P(10, 7) numbers without repeats, but then that's too much and I'm not sure how I can subtract the number's that N is not supposed to have. The other way I was thinking is the last 6 numbers can be chosen P(10, 6) without repeat's but then I'm not sure how I can choose the number of N. Can someone provide some help?
     
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  3. Nov 12, 2004 #2

    NateTG

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    How many single digit numbers with no repeated digits?
    What about two, three, four etc.?

    Also, it might be easier to figure out how many possible 7 digit numbers there are, and then subtract off the ones starting with 0 or 1.
     
  4. Nov 12, 2004 #3

    THA

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    It's regular 7-digit phone number, none of the 7 digit's can repeat. If that's not what you meant, then please clarify.

    That's what I was thinking, but I'm not really sure how to count the number's that start with 0 or 1. Would it be something like: P(10,7) - 2*P(10,6) ?
     
  5. Nov 12, 2004 #4

    Gokul43201

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    Here's what NateTG means : Forget for now that you have a 7 digit number and look at the following.

    How many ways of finding a 1-digit number ? That's easy : 10 ways. Any number from 0..9 can be used.
    How many 2-digit numbers without repeats ? Okay, I can put any of the10 numbers in the ten's place, and having used up this number, I can put any of the remaining 9 numbers in the unit's place. So there's 90 possibilities or P(10,2)
    In this 2-digit number, if the ten's digit had to be only one of 2..9, I can fill that spot in 8 ways, and the units spot in...still only 9 ways.

    And so on...
     
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