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Combinations problem

  1. Aug 16, 2014 #1
    Question: You are travelling to South America in two weeks. A friend bought you 5 watches and 6 pairs of sunglasses. You can bring at least 2 watches and at least 1 pair of sunglasses, and can only bring 4 items (so you don't lose all of them). How many combinations of watches and sunglasses can you have?

    This is my thought process:
    First, 5C2 (for the at least 2) = 10
    Second, 6C2 (to fill remaining last 2 "spots") = 15
    Multiply subsets = 150.

    5C3 (since it was at least 2) = 10
    6C1 (to fill in last spot) = 6
    Multiply subset = 60

    Add 150+60 = 210 possible combinations....

    is this correct? is my thought process right. with the limit of "4 items" i feel like i didn't need to add "at least 1 pair of sunglasses"

    what if the question instead stated, "you only want to bring 2 watches and 2 sunglasses...." then would it just be
    5C2 = 10
    6C2 = 15
    and then multiply? to get 150 combinations?

    thank you
     
  2. jcsd
  3. Aug 16, 2014 #2
    sorry if i didn't follow the preformatted thread
     
  4. Aug 16, 2014 #3

    Orodruin

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    It looks reasonable to me.

    However, you did add the information on "at least 1 pair of sunglasses". If not you would also have had the option of taking 4 watches.

    The following reasoning which gives another result would be wrong, can you tell why?
    I will bring at least two watches and one pair of sunglasses so let me start by picking these. That would give me 5C2 = 10 and 6C1 = 6 options, respectively. There now remains 8 objects of which I can pick one for 8C1 = 8 options. Multiplying together gives 10*6*8 = 480 combinations.
     
  5. Aug 16, 2014 #4

    haruspex

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  6. Aug 16, 2014 #5
    yeah sorry, i didn't mean to. didn't know where to ask until i started scoping around and found that this was the right place!!

    my apologies!
     
  7. Aug 16, 2014 #6
    ah okay so the 8C1 is to fill in the last "remaining" spot
     
  8. Aug 16, 2014 #7

    Orodruin

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    No, as I told you your first reasoning was correct. I added this as an example of how people get it wrong and asked if you can tell why it is wrong. (It is obviously double counting some combinations since the result was larger. But why does it overcount?)
     
  9. Aug 16, 2014 #8
    yes sorry i was referring to your example. and it would over count because now we might count in the "already chosen" watches and sunglasses
     
  10. Aug 16, 2014 #9

    Orodruin

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    Yes, some of the combinations where a particular watch is chosen in the last step will be equivalent to combinations where it was chosen in the first and so on. Therefore, that method overcounts.
     
  11. Aug 16, 2014 #10

    haruspex

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    As I pointed out on the other thread, it is not clear that you have to take four items. Do you?
     
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