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Combinations of n objects taken r at a time

  1. Jan 29, 2013 #1
    An office furniture manufacturer that makes modular storage files offer its customers two choices for the base and four choices for the top, and the modular storage files come in five different heights. The customer may choose any combination of the five different-sized modules so that the finished file has a base, a top, and one, two, three, four, five, or six storage modules. How many choices does the customer have if the completed file has four storage modules, a top, and base? The order in which the four modules are stacked is irrelevant.

    The best i could get was nCr=(n/r), (2/1)x(4/1)x((5/4)+(5/3)+(5/2)+(5/1)), and it is wrong. my teacher spent half an hour on this one question and he has no clue how to do it. This is basically my last resort.
     
  2. jcsd
  3. Jan 29, 2013 #2

    Simon Bridge

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    It's like the ice-cream flavors problem - only the number of choices for top and bottom flavor are not the same as for the middle ones. You've done that one before right?
    It would help to see your reasoning - but it looks like you did something to do with combinations out of 5 (but it looks like you got the formula mixed up) , times the 2 bases, times the 4 tops? The answer cannot be correct because it gives you a fraction in the answer - and you cannot have a fraction of a combination. In fact, the method you tried will usually give you a fraction. (BTW: do you know the correct answer?)

    In the description:
    You have a base (out of 2), a top (out of 4) so that's 8 right there... well done.
    The middle part has 4 modules in the middle, which the customer assembles out of 5 varieties.
    If the varieties are A,B,C,D,E, then one choice may be AABE, another may be ACCE.
    If order matters, then AABE is different from ABAE etc.

    Is the way forward clearer now?

    Note: it is easy to get mixed up about formulas, so try to look at what the formulas are trying to describe:
    http://www.mathsisfun.com/combinatorics/combinations-permutations.html
     
    Last edited: Jan 29, 2013
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