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Math Help - [SOLVED] Counting Problem - just a quick question with this problem

  1. #1
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    [SOLVED] Counting Problem - just a quick question with this problem

    Part a was this:
    Q: How many positive integers between 1000 and 9999 inclusive are divisible by 9?
    A: 9999 - 1000 + 1 = 9000

    This question is part c.
    Q: How many positive integers between 1000 and 9999 inclusive are distinct digit?
    __________________________________________________ ________________

    part c I do not understand.
    I have the answer as 9 x 9 x 8 x 7 = 4536. But I do not get why the 0 is not included in the 1st digit but the 0 is included in the 2nd?

    This is what was given by my instructor:
    1st digit is 9
    2nd digit is 10 - 1 = 9
    3rd is 10 - 2 = 8
    4th is 10 -3 = 7
    This is where he got 9 x 9 x 8 x 7.
    Last edited by Grillakis; April 13th 2009 at 04:41 PM.
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  2. #2
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    Why ?
    What is the wrong with your instructor`s answer ?
    1st digit - 1,2,3,4,5,6,7,8,9 (9digits)
    2nd digit - since we have selected 9 natural digits for the digit we have 8 more such digits but we can use zero for 2 nd digit . (therefore 9 digits again)
    3rd digit - we have 8 digits
    4th digit - we have 7 digits

    As you said if we have 0 for the 1st digit it will be less than 1000 ?
    won`t it be ?
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  3. #3
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    forget about it....I just figured it out.
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  4. #4
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    Thumbs up

    Quote Originally Posted by K A D C Dilshan View Post
    Why ?
    What is the wrong with your instructor`s answer ?
    1st digit - 1,2,3,4,5,6,7,8,9 (9digits)
    2nd digit - since we have selected 9 natural digits for the digit we have 8 more such digits but we can use zero for 2 nd digit . (therefore 9 digits again)
    3rd digit - we have 8 digits
    4th digit - we have 7 digits

    As you said if we have 0 for the 1st digit it will be less than 1000 ?
    won`t it be ?
    Think what happens if you select 0 for the 1st digit ...
    It will make your number less than 1000 .
    Can you understand it ?
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