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Counting problems

  1. Oct 4, 2009 #1
    1. The problem statement, all variables and given/known data

    how many 4 digit positive integers have at least one digit that is a 2 or a 3?

    2. Relevant equations

    - this is what I need -

    3. The attempt at a solution

    I cannot find the equation to this problem. Can someone give me a hand?
     
  2. jcsd
  3. Oct 4, 2009 #2

    jbunniii

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    Can you answer these questions?

    (1) How many 4-digit positive integers have NO digits that are 2 or 3?
    (2) How many 4-digit positive integers are there?
    (3) How can I use (1) and (2) to answer the original question?
     
  4. Oct 4, 2009 #3
    I have no clue of #1
     
  5. Oct 4, 2009 #4
    without using any formulas, I would just guess the answer is 2222... is that close?
     
  6. Oct 5, 2009 #5

    jbunniii

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    Let's start small.

    How many 1-digit positive integers have NO digits that are 2 or 3?
    How many 2-digit positive integers have NO digits that are 2 or 3?
    What's the general rule for an n-digit positive integer?
     
  7. Oct 5, 2009 #6

    jbunniii

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    By the way, notice that in order to have a four-digit number, the first digit has one more constraint. It can't be 2 or 3, but it also can't be ...?
     
  8. Oct 5, 2009 #7
    7
    49?
    7^n?
     
  9. Oct 5, 2009 #8
    Hm, so ...

    7
    8^(n-1) + 7?
     
  10. Oct 5, 2009 #9

    jbunniii

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    You're on the right track.

    For the FIRST digit you have 7 choices. For all the other digits you have 8 choices. So what's the general rule for n digits?
     
  11. Oct 5, 2009 #10

    jbunniii

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    Almost but not quite. Hint: it's not a "+", it's a ...?
     
  12. Oct 5, 2009 #11
    ahhh... [tex]8^{n-1} * 7[/tex] ?
     
  13. Oct 5, 2009 #12

    jbunniii

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    Yes. Do you understand why it's * and not +?

    So that answers my question (1).

    Now how about question (2)? This is much easier.

    Then question (3) is the key.
     
  14. Oct 5, 2009 #13
    9998 - 3584 = 6414?
     
  15. Oct 5, 2009 #14

    jbunniii

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    Close but not quite right.

    How many four-digit numbers are there? The first digit has to be 1-9, the other three digits can be anything.
     
  16. Oct 5, 2009 #15
    I'm not sure... why is it not 9998?
     
  17. Oct 5, 2009 #16

    jbunniii

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    Well, there are 9 choices for the first digit, 10 for the second digit, 10 for the third digit, and 10 for the fourth digit.

    So there are [itex]9 \times 10 \times 10 \times 10[/itex] possibilities in total. That's 9000, not 9998.

    If that's not clear, consider that you are excluding precisely the numbers 0000 through 0999. That's 1000 numbers excluded, out of 10000 possible combinations of digits, leaving 9000.

    Why are we excluding the numbers 0000 through 0999? Because written properly they are 0 through 999, which aren't 4-digit numbers!
     
  18. Oct 5, 2009 #17
    So why would 9990 for example not be valid? It is positive and uses 0 in the one's position. I assumed I could not use 0 but I could use 10, 100, 1000, etc.
     
  19. Oct 5, 2009 #18
    ooooooooooooooooooooooooooooooooooooooooo, so I have:

    9000 -3584 = 5416 that do not have a 2 or 3 ?
     
  20. Oct 5, 2009 #19

    jbunniii

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    Bingo!
     
  21. Oct 5, 2009 #20
    hurah! mucho gracias! Let's go to my new counting post! :p
     
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