How Can Students Be Arranged into Groups with Set Sizes?

AI Thread Summary
The discussion focuses on how to arrange 30 students into groups of specified sizes. For part a, the correct formula for arranging the students into groups of 5, 10, and 15 is confirmed as 30!/(5!10!15!), while the method of calculating combinations is discussed. In part b, it is clarified that when forming three groups of 10, the answer includes a division by 3! to account for identical groups, which avoids overcounting. The importance of context in determining whether groups are distinct or indistinct is emphasized, affecting how calculations are approached. Overall, understanding the nuances of group arrangement and the implications of group identity is crucial for accurate calculations.
Matejxx1
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Homework Statement


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. answer=\frac{30!}{5!*10!*15!}
b)we must have three group and all must have 10 students
answer= \frac{30!}{10!*10!*10!*3!}

Homework Equations


C^r_n = \frac{n!}{r!*(n-r)!}
V^r_n = \frac{n!}{(n-r)!}
P_n= {n!}

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

C^5_{30} = \frac{30!}{5!*(25)!}

and then I did the same for the other 2 groups
C^{10}_{25} = \frac{25!}{10!*(15)!}
C^{15}_{15} = \frac{15!}{15!*(0)!}
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)
C^{10}_{30} = \frac{30!}{10!*(20)!}
C^{10}_{20} = \frac{20!}{10!*(10)!}
C^{10}_{10} = \frac{10!}{10!*(0)!}
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
 
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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.
 
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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?
 
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.
 
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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?
 
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.
 
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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 ?
 
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.
 
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.
 
  • #10
RUber said:
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.
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 don't work on the same problem then the 3 factorial is not needed.
 
  • #11
TheMathNoob said:
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 don't work on the same problem then the 3 factorial is not needed.
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.
 
  • #12
haruspex said:
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.
mmm I just said subjective because it sounds cool jajajajaj, what is subjective?
 
  • #13
TheMathNoob said:
mmm I just said subjective because it sounds cool jajajajaj, what is subjective?
Subjective means it is a matter of opinion, as opposed to objective, meaning it is a matter of testable fact.
 
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