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Let me know how much time you took .. - thanks.. A teacher [#permalink] New post 28 Jun 2006, 02:51
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A teacher is assigning 6 students to one of three tasks. She will assign students in teams of at least 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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 [#permalink] New post 28 Jun 2006, 03:09
possible groups of teams
1,2,3 people per task-60 ways
2,2,2 people per task-90 ways
4,1,1 people per task-30 ways
SO (B)
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 [#permalink] New post 28 Jun 2006, 03:13
BG wrote:
possible groups of teams
1,2,3 people per task-60 ways
2,2,2 people per task-90 ways
4,1,1 people per task-30 ways
SO (B)



these are group formation ways. what abt team-assignment possibilities ??
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 [#permalink] New post 28 Jun 2006, 03:34
BG wrote:
possible groups of teams
1,2,3 people per task-60 ways
2,2,2 people per task-90 ways
4,1,1 people per task-30 ways
SO (B)


Agree with B.

Three tasks hence three teams. It cannot be greater than 3 as then one task will be assigned to more than one team.
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 [#permalink] New post 28 Jun 2006, 04:05
Not able to understand it, Please can someone explain how we got 60,90, and 30 values.
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 [#permalink] New post 28 Jun 2006, 19:43
humans wrote:
Not able to understand it, Please can someone explain how we got 60,90, and 30 values.


The first combination is 1, 2 , 3

One person in a team can be chosen in 6C1
2 persons from 5 left = 5C2
3 persons from 3 left = 3C3
Hence = 6C1*5C2*3C3 = 60

The second combination is 2, 2, 2
2 persons from 6 can be chosen in 6C2
2 persons from 4 can be chosen in 4C2
2 persons from 2 can be chosen in 2C2
Hence 6C2*4C2*2c2 = 90

The third combination is 4,1,1
4 persons from 6 can be chosen in 6C4
1 person from 2 can be chosen in 2C1
1 person from 1 can be chosen in 1C1
Hence 6C4*2C1*1C1 = 30
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 [#permalink] New post 28 Jun 2006, 19:51
one more for (B)

as explained above.
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 [#permalink] New post 28 Jun 2006, 20:21
Possible combinations of TEAMS TO TASKS (Not only teams):

6 Then 1* 3 = 3
1,5 Then 6C1 * 5C5 * 6 = 36
2,4 Then 6C2 * 4C4 * 6 = 90
3,3 then ((6C2*3C3)/2!) * 6 = 60
1,1,4 then 6C1 * 5C1 * 4C4 * 9 = 270
1,2,3 then 6C1 * 5C2 * 3C3 * 9 = 540
2,2,2 then ((6C2 * 4C2 * 2C2)/3!) * 9 = 135

So its B.
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 [#permalink] New post 29 Jun 2006, 00:10
I am not sure this is the correct method but got the same anwser. Just scanned the coices and saw which choices were divible by 6.
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 [#permalink] New post 29 Jun 2006, 04:31
jaynayak wrote:
humans wrote:
Not able to understand it, Please can someone explain how we got 60,90, and 30 values.


The first combination is 1, 2 , 3

One person in a team can be chosen in 6C1
2 persons from 5 left = 5C2
3 persons from 3 left = 3C3
Hence = 6C1*5C2*3C3 = 60

The second combination is 2, 2, 2
2 persons from 6 can be chosen in 6C2
2 persons from 4 can be chosen in 4C2
2 persons from 2 can be chosen in 2C2
Hence 6C2*4C2*2c2 = 90

The third combination is 4,1,1
4 persons from 6 can be chosen in 6C4
1 person from 2 can be chosen in 2C1
1 person from 1 can be chosen in 1C1
Hence 6C4*2C1*1C1 = 30


Thanks Jay for the wonderful explanation. :)
  [#permalink] 29 Jun 2006, 04:31
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