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Homework Help: Application of r-groups of n-objects

  1. Oct 20, 2016 #1


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    1. The problem statement, all variables and given/known data
    How many ways can the letters {"T", "O", "O", "T"} be arranged?
    How many groups of project team pairs can be formed from a pool of 4 students (2 math majors and 2 physics majors)?

    2. Relevant equations
    $$\text{outcomes} = \frac{n!}{n_1!n_2! \dots n_r!}$$

    3. The attempt at a solution
    In both cases, the computation is exactly the same $$\frac{4!}{2!2!} = 6$$ but the meaning of the result seems to be different. In the case of the letter arrangements, there are 6 un-ordered outcomes (meaning the order of the "T"s and "O"s doesn't matter because they are indistinguishable), but in the case of the project team pairs, there are 6 ordered outcomes and we need to divide by 2 to obtain the number of un-ordered outcomes. This seems contradictory. Same calculation, same result, but different meaning of the result. How is one suppose to determine which interpretation is valid? Is it as simple as saying that the two "O's and two "T"s are not distinguishable from each other (implying un-ordered result) but the students are (meaning ordered result)? I don't think so.

    If we labeled the letters, then [itex](T_1 T_2 O_1 O_2)[/itex] would be considered different from [itex](T_2 T_1 O_2 O_1)[/itex] but in the case of the project teams {[itex](M_1 M_2), (P_1 P_2)[/itex]} and {[itex](M_2 M_1), (P_2 P_1)[/itex]} would be considered the same. Wow! Maybe the difference is that these are completely different problems and thus need to be looked at differently?
    Last edited: Oct 20, 2016
  2. jcsd
  3. Oct 20, 2016 #2

    Simon Bridge

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    Different calculations are motivated by different needs - but they can still come out to the same number in the end. 2+2=4, so does 1+3.

    In your example - take a closer look:
    If Alice (A) and Bob (B) are physics (P) majors, while Carol (C) and Dave (D) are maths (M) majors; each person is, thus, characterized by two variables: (name, major) to make the set: {(A,P),(B,P),(C,M),(D,M)}, and your task is to count the number of pairs of people who happen to also have some majors.

    Then a pairing of Alice and Carol {A,C} is different from a paring of Alice and Dave {A,D} ... even though both sets are type {P,M}. This is because the requirement for pairs refers to the people, not their majors. But we notice that the pairing {A,C} is the same two people as {C,A} ... if it does not matter who comes first.

    OTOH: if you needed to know the possible combinations of majors instead, the calculation is different. Then {A,C} and {A,D} are the same combination of majors - even though they are different pairings of people.

    To know what matters, you need a careful reading of the problem description and some fluency with, in this case, English. It can help just to list all the possible pairs using all their characterizing variables then sort them according to the additional criteria.

    In TOOT, there is no intrinsic difference between the letter T that comes at the start and the letter T that comes at the end.
    Jumble them up and you cannot tell them apart... but it may be that the setup cares about where the letter came from - in which case, you need to label them appropriately. ie, T1,O2,O3,T4 ... then if you want to find the pairs of letter positions, you get the same calculation as above, with 1,2,3,4 playing the roles of A,B,C,D.

    There is no reason to individually identify the individual letters like we have to identify the individual people with the majors.
    Summary: the calculations are different because the problems are asking for different things.
  4. Oct 20, 2016 #3


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    I guess what I'd really like to know is what clue or clues to look for to determine if the result is "ordered" value or "un-ordered" value and if ordered, by what value to divide by to get the un-ordered result. Is the only way to work out a small problem in detail?
  5. Oct 22, 2016 #4


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    I found a video that clearly explains the differences. One thing I'm beginning to realize is that my conceptual viewpoint is flawed. I was trying to use calculus as an analogy as to how to approach problems. For example, we have a methodology for determining the partial derivative of a function. If you follow the rules of partial differentiation you get the partial derivative for your function and it means exactly the same thing in all cases. I was using the same mentality for combinatorics and have come to the conclusion that the analogy simply doesn't apply. The tools are much lower level and don't have any inherent meaning by themselves.
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