A teacher is assigning 6 students to one of three tasks. She

Question

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 has only one team assigned to it., the which of the following is/are possible combination(s) of teams to tasks?

1. 90
2. 60
3. 45
4. 30
5. 540

1. 90
2. 60
3. 45
4. 30
5. 540

durgesh79 · Answer

is there more than one correct answer ?...

iamcartic · Answer

Yes - read the stem carefully! :wink:

durgesh79 · Answer

we can divide the team in two ways 
1,1,4 or 1,2,3

ways for 1,1,4 = 6! / 4! = 30 
ways for 1,2,3 = 6! / (2! * 3!) = 60

if we take first case possbile configurations = 30*3! = 180 
if we take second case, possbile configuartion = 60*3! = 360

total ways 540

now the question language is a bit confusing.... IMO 30,60,90 and 540 can be possible ways....

sset009 · Answer

question

ways for 1,1,4 -
isnt it a case of choosing 4 people from a team of six irrespective of order, (coz ur accounting for the 3 teams when you multiply by 3! later)
and therefore shouldnt this be 6C4 = 6!/( 4! * 2 !) = 15?

and could you please expand on "ways for 1,2,3 = 6! / (2! * 3!) = 60"

thanx durgesh

bhushangiri · Answer

What about the possibility of 2,2,2 split ?

Also, i thought that 4,1,1 should have 3*6C4*2 = 90 ways.

You have 3 tasks. Let task 1 get 4 kids and task 2 and 3 get 1 kid each. 
Task 1 can get 4 kids out of 6 in 6C4 ways. The remain two kids can be assigned to the remaining 2 tasks in 2 ways. So total number of ways is 6C4*2.
But now you can have task 2 with 4 kids or task 3 with 4 kids. Each will get you 2*6C4 ways.

Therefore total ways of 4,1,1 split is 3*(6C4*2)=90.

durgesh79 · Answer

sset009 : if you take 6C4 the two people are left, these 2 people can be arranfed to two teams in 2 ways, so total number of ways will be 15*2 = 30

bhushangiri : Thanks for correcting me, yes (2,2,2) will also be an option

ways = 6! / (2! * 2! * 2!) = 90
possible configuration = 90 * 3! = 540

total possible configurations = 540 + (540 earlier) = 1080

regarding your comments about 1,1,4 combination... there are six kids numbered 1,2,3,4,5,6 three team A, B, C and three tasks X, Y, Z
number of ways 3 teams can do 3 tasks = 3! = 6 ways
AX, BY, CZ
AX, BZ, CY
AY, BX, CZ
AY, BZ, CX
AZ, BX, CY
AZ, BY, CX

Number of ways 6 people can be divided in 3 groups of 1,1,4 = 6C4 * 2 = 15*2 = 30

total number of ways = 30*6 = 180

durgesh79 · Answer

there is a formula for dividing n things, with p in one group, q in one group .... = n! / p! * q!

durgesh79 · Answer

oh i think i get it ...i'm making a mistake in (114) and (222) combinations

1,1,4 will be = 90  NOT 180  becuase order of first two team is not important 
2,2,2 will be = 90  NOT 540  becuase order of all teams is not important
1,2,3 will be = 360 as order is relevant.

judokan · Answer

can someone post OA and explanation? pls

bhushangiri · Answer

I am not sure i understand the principle behind (222) case. For this i am getting 540. Here is how.
We need to form 3 groups of 2 students each.

For the first two students we have 6C2 ways. For the next two students we have 4C2 ways and only 1 way for the last 2.
so 6C2 * 4C2 * 1 = 90

Now these 3 groups can be assigned to 3 tasks in 3*2*1 = 6 ways.

So total number of ways the tasks can be done = 90*6 = 540

Can you explain where i am taking order into account here ?

Ah ... I think i get it... tell me if my reasoning is same as yours..
When i write 6C2 * 4C2 * 1, i am already accounting for the possibility of any of the 6 kids working on task 1. and 2 of the remaining 4 for task 2. So i need not multiply by 6 after this point.

durgesh79 · Answer

exactly..... so kids number 1,2 getting task X, is already counted once in 90..so whether kids 1,2 are in team A, B or C doesnt matter.

iamcartic · Answer

Durgesh was right in his first reply - that is the OA!

bhushangiri · Answer

You mean second (or possibly third) reply where he included the corrected (1,1,4) and (2,2,2) cases and 540 is the final answer right ?

durgesh79 · Answer

:wink: technically in all cases the answer was right ... becuase the language of the quesation is not standatd GMAT ... it's looking for all possible configuration so more than one options are right...

1. 90 possible 
2. 60 possible
3. 45 - not possible in any configuartion..... 
4. 30 - possible 
5. 540 - possible

A teacher is assigning 6 students to one of three tasks. She

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