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Combinatorial question: permutation, binomial coefficient

  1. Mar 4, 2013 #1
    How many numbers of 6 digits which have exatctly the digit 1 (2 times), digit 2 (2 times), without zero, are there?
    The book post this solution: [tex] \frac{6!}{2!2!}*\binom{7}{2} + \frac{6!}{2!2!2!}*7= 4410[/tex],
    but i'm trying to find an explanation for this result.
     
  2. jcsd
  3. Mar 4, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    That is a weird way to calculate the number.

    Here is what I would do:

    There are ##6 \choose 2## ways to place the "1"s, and ##4 \choose 2## ways to place the "2"s afterwards. If you write both with factorials, a 4! cancels and you get ##\frac{6!}{2! 2! 2!}##. The remaining two digits have 7 options each (3...9), therefore the total number of digits is...

    Of course, you can split 7*7 in ##2 {7\choose 2} + 7##, but where is the point? The first part corresponds to the number of numbers where the two remaining digits are different, but I don't see a reason to consider them separately.
     
  4. Mar 4, 2013 #3
    It is [tex]\binom{6}{2}*\binom{4}{2}*7^2[/tex] by the general form of the product rule, where there are for any pairs of digits 3 cases, if I'm not wrong, thank you.
    I suppose to get this weird formula it calculate permutation with repetition and it brokes the problem in two sets.
     
  5. Mar 4, 2013 #4

    ssd

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    The 1st term corresponds to the case when the other two digits are different and the 2nd term corresponds to the case when another digit is repeated (three double digits).
     
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