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Combinations: Practice Question of the Week #5

This is a pretty difficult probability/integer properties hybrid, good luck!

In how many ways can 16 different gifts be divided among four children such that each child receives exactly four gifts?

A. 16^4

B. (4!)^4

C. (16!)/(4!)^4

D. (16!)/(4!)

E. 4^16


Update: Here’s the answer/explanation post!

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3 Responses to Combinations: Practice Question of the Week #5

  1. PavSwede October 6, 2011 at 2:37 am #

    The other way to solve this involves no math at all: The question gives you three pieces of data, presumably all important to solving the problem (16 gifts, 4 children, 4 gifts for each child). The only answer that includes all three pieces of data in it is C. Sometimes multiple choice questions are a matter of logic and not knowing the math.

    • Chris Lele
      Chris October 6, 2011 at 3:46 pm #

      Interesting… but I don’t think that method is necessarily valid and applies to all cases. For instance, if I removed the constraint for 4 gifts per child, then I would have a far more difficult problem, not, as your logic would suggest, an easier problem with fewer moving parts, so to speak.

      At the same time, I think often you can work through a problem based on logic. However, with this problem I think your assumption, though it got you the right answer, would not pertain in other cases.

  2. Nirav Aga June 28, 2011 at 1:25 pm #

    The answer is C.
    Firstly for selecting 4 gifts for a child, the number of possible combinations is 16C4.
    For the next child, the possible selections are 12C4 and for the third, it’s 8C4.
    On multiplying all the possible selections, the value obtained is 16/(4!)^4.
    That gives the final solutions.
    Also, as the arrangement of the gifts is not important, the answer cannot be 16!/4!.

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