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Integers 0-9999999 containing 2 3's and 2 5's

  1. Apr 21, 2009 #1
    1. The problem statement, all variables and given/known data
    How many integers from 0-9,999,999 have exactly two 3's and two 5's as digits.


    2. Relevant equations
    I'm not really sure...


    3. The attempt at a solution
    The answer is 107520, if I'm not mistaken. I made a program to count it up for me, so I'm fairly sure that that is the correct answer. I'm just trying to figure out how to do it manually now.

    So I know there are 7 digits, 4 of which are 3, 3, 5, 5, and the other 3 are 0,1,2,4,6,7,8,9.
    So the remaining 3 digits each have 8 possibilities. So there are 8^3 different combos for the other 3 digits. But then all of the digits can be rearranged, so I thought it would be 7! * 8^3. That was not even close to correct.
     
  2. jcsd
  3. Apr 21, 2009 #2
    Well, each of the 7 digits can be any digit from 0...9. So we have the numbers

    0 0 0 0 0 0 0
    0 0 0 0 0 0 1
    ...
    9 9 9 9 9 9 8
    9 9 9 9 9 9 9

    Clearly, there are 10^7, or 10 million, of these.

    So, you have to pick 2 from 7 to call "3". Then you have to pick 2 from the remaining 5 places to call "5". Then you can choose any of 8 digits to fill the last three spots.

    If you use this information, you WILL get the right answer. If I were to do the next step, you would get the answer. That's how close you are now. Focus on the logic I used and understand that, as that's what's most important.
     
  4. Apr 21, 2009 #3

    Dick

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    Science Advisor
    Homework Helper

    Not all rearrangements give rise to different numbers. You are over counting. First pick two positions out of the 7 to place the 3's. How many ways to do that?
     
  5. Apr 21, 2009 #4
    Ooooh ok, I see.
    so its 7 choose 2 * 5 choose 2 * 8^3
    Thank you so much!
     
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