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Permutation & combination

  1. Sep 24, 2015 #1
    1. The problem statement, all variables and given/known data
    There are 30 students in a class. In how many ways can we arrange them if :
    a)we must have three group, group one must have 5 students , group two 10 students and group three 15 students. [tex]answer=\frac{30!}{5!*10!*15!}[/tex]
    b)we must have three group and all must have 10 students
    [tex]answer= \frac{30!}{10!*10!*10!*3!}[/tex]

    2. Relevant equations
    [tex]C^r_n = \frac{n!}{r!*(n-r)!}[/tex]
    [tex]V^r_n = \frac{n!}{(n-r)!}[/tex]
    [tex]P_n= {n!}[/tex]
    3. The attempt at a solution
    a)
    Okay so I'm not sure if the way I solved a) is correct, as I didn't get the exact same result as I should have .
    what I tried doing was: to first pick 5 students out of those 30 so I got

    [tex]C^5_{30} = \frac{30!}{5!*(25)!}[/tex]

    and then I did the same for the other 2 groups
    [tex]C^{10}_{25} = \frac{25!}{10!*(15)!}[/tex]
    [tex]C^{15}_{15} = \frac{15!}{15!*(0)!}[/tex]
    and then I multiplied all the 3 combinations together and got
    4.66*1011
    which is the right answer but not the same one as in the textbook.

    b) I tried doing the same thing as with a)
    [tex]C^{10}_{30} = \frac{30!}{10!*(20)!}[/tex]
    [tex]C^{10}_{20} = \frac{20!}{10!*(10)!}[/tex]
    [tex]C^{10}_{10} = \frac{10!}{10!*(0)!}[/tex]
    and the result comes out completely wrong
    I am wondering now if somebody could help me figure a) out and kinda of point me in the right direction with b)
    thanks
     
  2. jcsd
  3. Sep 24, 2015 #2

    RUber

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    For part a, you have the right answer. Write the 3 terms next to each other and cancel out factorials that appear on both the top and bottom of the multiplication.
    For part b, since the 3 groups are all the same size, there is a chance that Groups A, B, and C in one of your listed options is the same as group B, A, C or C, A, B, etc. Since you have to remove the duplications, that is why you see the 3! in the denominator of the answer you posted. There are 3! ways to permute the 3 groups to have the same members in each of the 3 groups.

    Note that you would not need to do that if the groups were in some way different...i.e. one group of 10 goes to the store, one group of 10 stays at work, and one group of 10 gets the day off.
    Other than that, you have the right method.
     
  4. Sep 24, 2015 #3
    Thank you for your answer.
    So if I understood you correctly I forgot to take into account that those 10 students who are in group A could also be in group B and C( and B in A and C , . . .). I can see now why that would cause a problem.
    So in general if you have more groups of the same size you must also take into account the duplicates?
    Also mind if I ask another question which is closely related to this one?
     
  5. Sep 24, 2015 #4

    RUber

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    I don't mind, but you often get faster responses (other helpers) if you post a new thread. Many people browse the unanswered posts first.
     
  6. Sep 24, 2015 #5
    7it's basically the same thing
    you have to arrange 10 people in 2 different groups (a group has to have at least 3 people in it )
    so I did this one like this
    C310*C410*C510*C610*C710
    So if I go with what you said about making sure there aren't any duplicates I should also divide this whole thing by 2! because the C77 can contain the same combinations as C710?
     
  7. Sep 24, 2015 #6

    RUber

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    So, do all 10 people have to be used?
    If so, I would add you combinations.

    10C3 for 3 in one group and 7 in another.
    If the groups are indistinct except for size, you notice that 10C3 is the same as 10C7.
    If the groups are distict like one group goes home and the other stays at work, then 10C7 would imply 7 go home whereas 10C3 would imply 3 go home.
     
  8. Sep 25, 2015 #7
    Yes sorry. What I meant to write was "plus" not "times" that's how I wrote it in my homework.
    So if the 2 groups are indistinct we must divide by the number of those groups factorial? and if thhey are distinct we just leave them as they are ?
     
  9. Sep 25, 2015 #8

    RUber

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    Right.
    If they are distinct, then you have: Groups like (3A, 7B), ( 4A, 6B), (5A,5B), (6A, 4B), (7A, 3B).
    If they are indistinct, then 10C7 duplicates 10C3, and 10C6 duplicates 10C4, and you divide the 10C5 by 2! since there would be inherent duplication between the two groups of 5.
     
  10. Sep 25, 2015 #9

    haruspex

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    For what it's worth, you might find it slightly easier to start with all possible splits into two groups and subtract the 10-0, 9-1 and 8-2 cases.
     
  11. Sep 25, 2015 #10
    Hi Ruber, I am also a student who is currently learning probability and to me the 3! is very subjective to the context. For example, we could say that A is UC santa cruz, B is UC Berkeley, and C is UC Davis. In this context, 10 students could work in Santa Cruz or Berkeley or Davis and each one of them would be a different arrangement. But for example in this case if all those students work on the same problem then I agree that we have to take into account the 3 factorial but if they dont work on the same problem then the 3 factorial is not needed.
     
  12. Sep 26, 2015 #11

    haruspex

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    That doesn't make it subjective; it makes it context sensitive. Ideally, the question would make it clear whether the groups are considered distinct. With no indication, it is probably safer to assume they are not.
     
  13. Sep 26, 2015 #12
    mmm I just said subjective because it sounds cool jajajajaj, what is subjective?
     
  14. Sep 26, 2015 #13

    haruspex

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    Subjective means it is a matter of opinion, as opposed to objective, meaning it is a matter of testable fact.
     
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