Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 29 Jun 2016, 15:38

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Let me know how much time you took .. - thanks.. A teacher

Author Message
Senior Manager
Joined: 20 Feb 2006
Posts: 331
Followers: 1

Kudos [?]: 55 [0], given: 0

Let me know how much time you took .. - thanks.. A teacher [#permalink]

### Show Tags

28 Jun 2006, 03:51
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Let me know how much time you took .. - thanks..

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
Director
Joined: 13 Nov 2003
Posts: 790
Location: BULGARIA
Followers: 1

Kudos [?]: 43 [0], given: 0

### Show Tags

28 Jun 2006, 04:09
possible groups of teams
SO (B)
Senior Manager
Joined: 20 Feb 2006
Posts: 331
Followers: 1

Kudos [?]: 55 [0], given: 0

### Show Tags

28 Jun 2006, 04:13
BG wrote:
possible groups of teams
SO (B)

these are group formation ways. what abt team-assignment possibilities ??
SVP
Joined: 30 Mar 2006
Posts: 1737
Followers: 1

Kudos [?]: 65 [0], given: 0

### Show Tags

28 Jun 2006, 04:34
BG wrote:
possible groups of teams
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.
Manager
Joined: 12 Apr 2006
Posts: 218
Location: India
Followers: 1

Kudos [?]: 25 [0], given: 17

### Show Tags

28 Jun 2006, 05:05
Not able to understand it, Please can someone explain how we got 60,90, and 30 values.
SVP
Joined: 30 Mar 2006
Posts: 1737
Followers: 1

Kudos [?]: 65 [0], given: 0

### Show Tags

28 Jun 2006, 20:43
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
Senior Manager
Joined: 07 Jul 2005
Posts: 404
Location: Sunnyvale, CA
Followers: 2

Kudos [?]: 9 [0], given: 0

### Show Tags

28 Jun 2006, 20:51
one more for (B)

as explained above.
CEO
Joined: 20 Nov 2005
Posts: 2911
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 22

Kudos [?]: 208 [0], given: 0

### Show Tags

28 Jun 2006, 21:21
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.
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Manager
Joined: 31 Mar 2006
Posts: 162
Followers: 1

Kudos [?]: 17 [0], given: 0

### Show Tags

29 Jun 2006, 01:10
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.
Manager
Joined: 12 Apr 2006
Posts: 218
Location: India
Followers: 1

Kudos [?]: 25 [0], given: 17

### Show Tags

29 Jun 2006, 05:31
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.
Display posts from previous: Sort by