1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Permutation or Combination?

  1. Oct 3, 2011 #1


    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    The integers 1 through 6 appear on the six faces of a cube, one on each face. If three such cubes are rolled, what is the probability that the sum of the numbers on the top faces is 17 or 18?

    2. Relevant equations

    Probability = # of desired outcomes / total # of outcomes

    The total number of outcomes is 6*6*6 = 216.

    The number of desired outcomes depends. There could be 2, or 4.

    This is indisputable:

    6, 6, 6 => 18

    BUT ...

    6, 6, 5 => 17
    6, 5, 6 => 17
    5, 6, 6 => 17

    are the above sets distinguishable? Does AAB = BAA (combination)? Or is this a permutation problem, where AAB ≠ BAA?

    3. The attempt at a solution

    If this is a combination problem, the answer is 1/108. There is one way to get 18, and one way to get 17. If this is a permutation problem, the answer is 1/54 (there are 3 unique ways to get 17 and one way to get 18).

    Which one is correct? I think this question just screams AMBIGUOUS, but my teacher doesn't agree with me. We were covering permutations and combinations today in my Pre-Calc class, and my teacher keeps insisting the answer is 1/54, but I can see an argument for both sides.

    The answer key also says 1/54, but I think that's wrong. It should be both 1/108 and 1/54 given the open-ended nature of this question (as it is currently worded).
  2. jcsd
  3. Oct 3, 2011 #2
    It might help to think about the 216 possibilities, since you seem to accept that number without concern. Are 6, 6, 5 and 6, 5, 6 and 5, 6, 6 each on the list of 216 possibilities? If they are, then you must count them all in order to have the correct probability of occurrence.

    If it is not immediately apparent to you whether they are all on the list of 216, try examining two dice. That's 36 outcomes. Try tabling them, and see if 5+6 and 6+5 are in the table of 36 outcomes.
  4. Oct 3, 2011 #3


    User Avatar
    Gold Member

    I don't understand your analogy. If you were to throw 2 dice at once ... how can you distinguish between a 5,6 and a 6,5?

    As far as you can tell a 5,6 and a 6,5 are effectively the same things.

    If you were to throw the 2 dice separately, sure, you can tell between a 5,6 and a 6,5.

    The original question fails to make that distinction. What happens if I just randomly toss 3 dice. How in the world am I supposed to tell the difference between a 5,6,6 and a 6,5,5? There are still 216 possibilities.
  5. Oct 3, 2011 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    For ordinary (classical) dice, they are identical and not distinguished, but are *distinguishable* in principle. For example, we could use one red, one green and one blue die but which are otherwise exactly the same. Thus, it makes sense to speak of die 1, die 2 and die 3, but to ignore the numbers when they don't matter. They DO matter when counting the possibilities. There is nothing at all ambiguous about such situations.

    Strangely, if you had "quantum" dice you could not--even in principle--distinguish between them, and in that case the counting would be very different. If you had Bose-Einstein dice the number of outcomes giving a total of 17 would be 1; if you had Fermi-Dirac dice, you could not get a total of 17 or 18 at all because that would require at least two dice to have the same number. Anyway, maybe I should not have mentioned this at all; I hope it does not confuse you. When you see dice-tossing problems in normal probability courses it invariable refers to dice that we could distinguish between if we wanted to.

  6. Oct 4, 2011 #5
    Table the 36 possibilities of 2 dice.
  7. Oct 4, 2011 #6


    User Avatar
    Gold Member

    Well, that information would have been useful if it were presented in the original problem.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook