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Combination rule

  1. Sep 3, 2013 #1
    1. The problem statement, all variables and given/known data


    An accounting professional association is considering six hotels as possible sites for their next two meetings in 2001 and 2002. In how many ways can the association select the hotels...
    a) if the two meetings may be held at the same hotel?
    b)if the two meetings may not be held at the same hotel?
    2. Relevant equations



    3. The attempt at a solution
    I found the answers by drawing out a tree... But i was wondering if there was a mathematical way using the counting rules i.e permutation/combinations rules etc. I was trying for a while but couldn't find one
     
  2. jcsd
  3. Sep 3, 2013 #2

    Zondrina

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    For part a), there are six hotels and only two meetings to consider. So you only have to concern yourself with re-arranging the possibilities of the six hotels.

    ##{n \choose k}##

    Does this look familiar to you? How about part b)?
     
    Last edited: Sep 3, 2013
  4. Sep 3, 2013 #3

    CompuChip

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    Actually combinations will not work for a as they exclude the possibility of the hotels being the same in both years AND, more importantly, they imply the order doesn't matter while actually it is important here because the two years are not interchangeable (e.g. selecting hotel 1 then 2 is a different situation from selecting hotel 2 then 1).

    The answer is actually simpler: there are 6 possible choices in year 1 and for each of those there are 6 choices in year two. How many does that give you in total? (If you look back at your tree you may see that this is basically the same calculation except without drawing the tree).

    You can apply the same approach to b).
     
  5. Sep 3, 2013 #4
    OH okay i get it. Ty for the explanation guys ^^
     
    Last edited: Sep 3, 2013
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