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Homework Help: How many numbers betweeen 0 and 9999 have 2,5 and 8 as digitis?

  1. Jun 1, 2010 #1
    1. The problem statement, all variables and given/known data
    How many numbers betweeen 0 and 9999 have 2,5 and 8 as digitis?


    3. The attempt at a solution
    I have two attempts, each witha different result
    The first one:
    We need to arrange 4 numbers in a row such that three of the numbers are 2, 5 and 8. There are 10 options for selecting the last number and 4!=24 different arangements for a total of 10*24 = 240.

    The seconed one:
    In each block of 1000 numbers there are 6 combiantions. x258, x285...
    We can take the seven blocks that don't start with 2 5 or 8, for 42 numbers.

    Next we look at one block x--- where x is 2 5 or 8. for our purposes the blocks are symetrical so we don't need to look at them individualy.
    Lets say we take the block that starts with 2
    then the block 258x gives us 10
    then the block 285x gives us 10
    then the block 2x58 gives us 9
    then the block 2x85 gives us 9
    then the block 25x8 gives us 8
    then the block 28x5 gives us 8
    for a total of 2*(10+9+8)=52

    We have three such blocks for a total of 2*52=156.
    And adding the other 42 we get 198
    Whats wrong here?
    Thanks
    Tal
     
  2. jcsd
  3. Jun 1, 2010 #2

    benorin

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    I think you've counted some numbers twice.
     
  4. Jun 1, 2010 #3
    In both cases?
     
  5. Jun 1, 2010 #4
    English isn't my native language. What is a degenerate case?
     
  6. Jun 1, 2010 #5

    Hurkyl

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    Sorry, I deleted my post after I realized you have (I think) worse problems than the one I mentioned.


    A degenerate case is... well, it's hard to describe in words. It often amounts to something unusually special happening. e.g.

    Often, when working with sets of N objects, the N=0 case is degenerate.

    You are doing something in geometry with pairs of lines. The case when they are parallel is often degenerate. The case when they are equal even moreso.

    You are counting the number of ways to arrange a collection of objects. When two of them are indistinguishable, that is often degenerate.
     
  7. Jun 2, 2010 #6
    Ok How aboyt this way:
    We have 7 unimportant numbers {0,1,3,4,6,7,9,}. to make a four digit number with 2,5,8 and one of them we have 4!*7 options, and since each number is distinc there is no repetition.
    Looking at the other three we have 4!*3, but we need to divide by half because 2258 gets counted twice as does 2582. So we have $1*7+4!*7/2 = 204.
    I don't like that every attempt gives me a different awnser.
    ANyone know which one is correct, if at all? and what is the problem with the rest of them?
     
  8. Jun 2, 2010 #7

    Dick

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    204 is correct, and so is your reasoning.
     
  9. Jun 2, 2010 #8
    Hey Dick,
    Thans,
    Can you tell me where I went wrong with this one, I cant figure it out:
    The seconed one:
    In each block of 1000 numbers there are 6 combiantions. x258, x285...
    We can take the seven blocks that don't start with 2 5 or 8, for 42 numbers.

    Next we look at one block x--- where x is 2 5 or 8. for our purposes the blocks are symetrical so we don't need to look at them individualy.
    Lets say we take the block that starts with 2
    then the block 258x gives us 10
    then the block 285x gives us 10
    then the block 2x58 gives us 9
    then the block 2x85 gives us 9
    then the block 25x8 gives us 8
    then the block 28x5 gives us 8
    for a total of 2*(10+9+8)=52

    We have three such blocks for a total of 2*52=156.
    And adding the other 42 we get 198
     
  10. Jun 2, 2010 #9

    Dick

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    Doh. 2*(10+9+8)=54. Not 52. So you get 3*54+42=204. :)
     
    Last edited: Jun 2, 2010
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