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Combinations

by Vemuri » Tue Mar 17, 2009 6:18 pm
A teacher is assigning 6 students to one of three tasks. She will assign students in teams of atleast one student, and all students will be assigned to teams. If each task will have exactly one team assigned to it, then which of the following are possible combinations of teams to tasks?
I. 90
II. 60
III. 45

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II and III
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Re: Combinations

by Morgoth » Wed Mar 18, 2009 1:37 am
Vemuri wrote:A teacher is assigning 6 students to one of three tasks. She will assign students in teams of atleast one student, and all students will be assigned to teams. If each task will have exactly one team assigned to it, then which of the following are possible combinations of teams to tasks?
I. 90
II. 60
III. 45

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II and III
Since there are 3 tasks and every student needs to be assigned to a team, only 3 teams can be formed.

arrange 6 students in 3 teams

CASE I

2 2 2

6C2 * 4C2 * 2C2 = 15*6*1 = 90

CASE II

4 1 1

6C4 * 2C1 * 1C1 = 15*2 = 30

CASE III

3 2 1

6C3 * 3C2 * 1C1 = 20*3 = 60



Hence only 60 and 90 are the two possible answers. Thus, B.

OA?
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by avenus » Wed Mar 18, 2009 5:08 am
Hi Morgoth,

I believe there's something you've not taken into account: there are three different tasks. Once you have the three teams, you can still assign the tasks in 3! different ways, can't you?

This is the way I see it:


Split it in two steps.

a) Calculate the number of different teams that could be formed without taking order into account
b) Multiply the result obtained in a) by 3!, since we can assign one of three tasks to each team

Morgoth wrote:CASE I

2 2 2

6C2 * 4C2 * 2C2 = 15*6*1 = 90
a) Number of different teams is 6C2*4C2*2C2/3!
Note that if you leave it at 6C2*4C2*2C2, you will be counting each possible combination several (3!) times, since there are three teams with an equal number of members


b)3! * 6C2*4C2*2C2/3!

So, in this case, we agree on the result, although I suspect it's just coincidence

Morgoth wrote: CASE II

4 1 1

6C4 * 2C1 * 1C1 = 15*2 = 30

a) Number of different teams is 6C4*2C1*1C1/2!
Note that without the 2! on the denominator, we would we counting cases such as
ABCD E F and ABCD F E as different, which is not what we want
There are 2 teams with an equal number of members

b) 3!*6C4*2C1*1C1/2! = 3*6C4*2C1*1C1 = 3*30 = 90

This is different from what you get...

Morgoth wrote: CASE III

3 2 1

6C3 * 3C2 * 1C1 = 20*3 = 60
a) Number of different teams is 6C3*3C2*1C1
Note that in this case, unlike cases I and II, teams counted in the formula above will all be different, since there are no teams with an equal number of members.

b) As usual, 3!*6C3*3C2*1C1 = 60*3! = 360



Also different from your result...


Let me know what you think
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by Morgoth » Wed Mar 18, 2009 7:12 am
avenus wrote:Hi Morgoth,

I believe there's something you've not taken into account: there are three different tasks. Once you have the three teams, you can still assign the tasks in 3! different ways, can't you?

This is the way I see it:


Split it in two steps.

a) Calculate the number of different teams that could be formed without taking order into account
b) Multiply the result obtained in a) by 3!, since we can assign one of three tasks to each team

Morgoth wrote:CASE I

2 2 2

6C2 * 4C2 * 2C2 = 15*6*1 = 90
a) Number of different teams is 6C2*4C2*2C2/3!
Note that if you leave it at 6C2*4C2*2C2, you will be counting each possible combination several (3!) times, since there are three teams with an equal number of members


b)3! * 6C2*4C2*2C2/3!

So, in this case, we agree on the result, although I suspect it's just coincidence

Morgoth wrote: CASE II

4 1 1

6C4 * 2C1 * 1C1 = 15*2 = 30

a) Number of different teams is 6C4*2C1*1C1/2!
Note that without the 2! on the denominator, we would we counting cases such as
ABCD E F and ABCD F E as different, which is not what we want
There are 2 teams with an equal number of members

b) 3!*6C4*2C1*1C1/2! = 3*6C4*2C1*1C1 = 3*30 = 90

This is different from what you get...

Morgoth wrote: CASE III

3 2 1

6C3 * 3C2 * 1C1 = 20*3 = 60
a) Number of different teams is 6C3*3C2*1C1
Note that in this case, unlike cases I and II, teams counted in the formula above will all be different, since there are no teams with an equal number of members.

b) As usual, 3!*6C3*3C2*1C1 = 60*3! = 360



Also different from your result...


Let me know what you think
Firstly, the question is of combination not permutation

Secondly, I guess you are misreading the question, the question is only asking about number of ways you can form 3 teams out of 6 students.

I think you cannot multiply the number of selections * number of arrangements. That does not make any sense logically.

Let me know what do you think.
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by lilu » Wed Mar 18, 2009 10:59 am
I also got B:

Possible combinations:
1. 2+2+2 --> 6!/2!*2!*2!=90
2. 4+1+1 --> 6!/4!*1!*1!=30
3. 3+2+1 --> 6!/3!*2!*1!=60

So, II or B
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by avenus » Wed Mar 18, 2009 5:28 pm
Firstly, the question is of combination not permutation

Secondly, I guess you are misreading the question, the question is only asking about number of ways you can form 3 teams out of 6 students.

I think you cannot multiply the number of selections * number of arrangements. That does not make any sense logically.

Let me know what do you think.
I might be misinterpreting the question but the wording is certainly poor and ambiguous.
Assuming they only ask about the number of ways in which 3 teams can be formed with six students, I still don't see it the way you do. For instance, for case II:
Morgoth wrote: CASE II

4 1 1

6C4 * 2C1 * 1C1 = 15*2 = 30
Once the four-member team has been determined, so have the other two, since there are two people left for two one-person teams. We're then looking for the number of ways in which a team of four people can be formed out of six people, i.e.

6C4 = 15

No need to multiply by 2.... Where do you get the other 15 from??
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by Morgoth » Fri Mar 20, 2009 4:27 am
avenus wrote:
Morgoth wrote: CASE II

4 1 1

6C4 * 2C1 * 1C1 = 15*2 = 30
Once the four-member team has been determined, so have the other two, since there are two people left for two one-person teams. We're then looking for the number of ways in which a team of four people can be formed out of six people, i.e.

6C4 = 15

No need to multiply by 2.... Where do you get the other 15 from??

you have to form 3 teams out of 6 students

4 1 1

means


out of 6 students, 4 students are chosen = 6C4 = 15 ways

out of remaining 2, 1 student is chosen = 2C1 = 2 ways

out of remaining 1, 1 students is chosen = 1C1 = 1 way

total ways 3 teams could be formed of 4, 1 , 1 combination is

15*2*1 = 30 ways

Hope this helps.
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by avenus » Fri Mar 20, 2009 7:45 am
you have to form 3 teams out of 6 students
4 1 1
means
out of 6 students, 4 students are chosen = 6C4 = 15 ways
out of remaining 2, 1 student is chosen = 2C1 = 2 ways
out of remaining 1, 1 students is chosen = 1C1 = 1 way
total ways 3 teams could be formed of 4, 1 , 1 combination is
15*2*1 = 30 ways
Hope this helps.
Let's see. You're just repeating the same without commenting on what I added on my last post. You're counting each team twice. Check the following example:

Let's suppose (for the sake ot simplicity) we have four people instead of six and we want to form three teams:
2 1 1
Applying your rationale (and correct me if I'm wrong):
4C2*2C1*1C1 = 12

I say that's six too much. Once again, if you've selected the first two people for the two-person team, there's only one way of assigning the other two (since we are not considering order): each one goes to a one-person team.
Imagine we have ABCD. We choose the members of two-team people:
AB
Since we're looking for a 2 1 1 arrangement:
AB C D
and that's it. By multiplying 4C2 by 2C1 you are actually counting
AB C D and AB D C
as different but they aren't. If you do consider them as different, you would also have to consider
C AB D, C D AB, etc... and then the result would be 3!*4C2

The possible combinations for the 2 1 1:

AB C D
AC B D
AD B C
BC A D
BD A C
CD A B

So that's 6, not 12 as your approach would yield....
Hope this helps pull you back into the argument :)
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