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Homework Help: Combinations/permutations help

  1. Sep 18, 2010 #1
    1. The problem statement, all variables and given/known data

    How many (natural) numbers less than 100 contain a 3? (Note: 13, 35 and 73 all
    contain a 3 but 42, 65 and 88 do not).

    3. The attempt at a solution

    Of course I know that the numbers containing a 3 including 10 numbers starting with a 3 (30, . . . 39),and 10 numbers ending in a 3 (3, 13, . . . , 93), with 33 being counted twice, so a total of 19 numbers. I've found this by counting. But is there a quick systematic way of obtaining this answer using combinations/permutations etc? Unfortunently my knowledge of combinatorics is very poor, so I appreciate any help.

    Between 10 to 100 there are 98 2-digit numbers that can possibly contain a 3 in the 1's or 10's positions... I'm stuck here.
  2. jcsd
  3. Sep 19, 2010 #2
    Re: Combinations

    Not every combinatorics problem involves permutations or combinations.

    Think of it this way: You want to count the numbers of the form xy where x or y is a 3.

    It would be easy to overcount, so be careful.
  4. Sep 19, 2010 #3
    Re: Combinations

    The only count would be 33. So other than counting, there are no easy systematic ways of doing this?
  5. Sep 19, 2010 #4


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    Re: Combinations

    How many numbers are there where x is a 3?

    How many numbers are there where y is a 3?
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