Author 
Message 
Manager
Joined: 21 May 2011
Posts: 240

Problem solving  3 groups of 3 [#permalink]
Show Tags
19 Jul 2011, 05:20
Question Stats:
20% (00:00) correct
80% (02:26) wrong based on 6 sessions
HideShow timer Statistics
In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?
280 1,260 1,680 2,520 3,360
I am looking for an explanation, not the answer.



VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1325

Re: Problem solving  3 groups of 3 [#permalink]
Show Tags
19 Jul 2011, 09:09
from 9 people any 3 can be selected.hence there will be 3 such teams = 9c3/3. from 6 people any 3 can be selected.hence there will be 2 such teams = 6c3/2 from 3 people any 3 can be selected.hence there will be 1 such team = 3c3/1 thus 9c3 * 6c3 * 3c3/ 3*2*1 = 210.
_________________
Visit  http://www.sustainablesphere.com/ Promote Green Business,Sustainable Living and Green Earth !!



Intern
Joined: 18 Jul 2011
Posts: 48

Re: Problem solving  3 groups of 3 [#permalink]
Show Tags
19 Jul 2011, 12:10
Think of it like an experiment: 1. Select 3 people from 9 at random 2. Select another 3 people from the 6 remaining people 3. Select another 3 people for the 3 remaining people
The first thing to notice is that steps 1 and 2 completely determine step 3. The second thing to notice is that since we're selecting teams, order doesn't matter, so we should use combinations 9C3 = (9*8*7)/(3*2*1) =168 6C3 = (6*5*4)/(3*2*1) = 20
By the fundamental counting principle, we should multiply these two numbers to see how many outcomes our experiment has 168*20 = 3,460
BenchPrepGURU



Intern
Joined: 18 Jul 2011
Posts: 48

Re: Problem solving  3 groups of 3 [#permalink]
Show Tags
19 Jul 2011, 12:11
calculation error!!!!!
168*20 = 3,360
BenchPrepGURU



Intern
Joined: 14 Jun 2011
Posts: 2

Re: Problem solving  3 groups of 3 [#permalink]
Show Tags
19 Jul 2011, 13:23
Solution: The first group 9C3=84 The second group 6C3=20 And the last group 3C3=1 In total have 84∙20∙1=1680 ways, but, exist repetitions in ways, then: ( 9C3∙ 6C3∙ 3C3)/3!=1680/6=280



Manager
Joined: 21 May 2011
Posts: 240

Re: Problem solving  3 groups of 3 [#permalink]
Show Tags
22 Jul 2011, 09:28
BenchPrepGURU,
Just wanted to point out some mistakes in your approach so it doesn't mislead someone.
First, 9C3 = (9*8*7)/(3*2*1) = 84 not 168 6C3 = (6*5*4)/(3*2*1) = 20
Second, 1/3rd of the combinations will be repetitions in 3 groups of 3 and 1/2 of the combinations will be repetitions in 2 groups of 3. So you should divided the result be 3 and 2 respectively arriving at 280.



Senior Manager
Joined: 05 Jul 2010
Posts: 355

Re: Problem solving  3 groups of 3 [#permalink]
Show Tags
22 Jul 2011, 18:45
There are multiple ways to solve this question.
1. Anagram: Let's focus on how to use this method, so that you can reuse it. Solution: = 9!/ (3!*3!*3!) = 1680
2. Direct and more reliable solution is to just visualize one person at a time. Solution: = (ways to select 3 out of 9) * (ways to select 3 our of remaining 6) * (ways to select 3 out of remaining 3) = 9C3*6C3*3C3 = [9!/(6!*3!)] *[6!/(3!*3!)]*1 = 9!/(3!*3!*3!) = 1680
Ans: "C"
 I made a mistake early on. I assumed they are 3 distinct group and hence thought order mattered. But this is NOT indicated in the question and MY ASSUMPTION WAS WRONG!
Hence the ans needs to divided by 3! (the number of ways the 3 groups can be rearranged). Hence 1680/3!=280. Ans: A
Thanks manishgeorge for pointing it out.
Hope this helps!
Last edited by abhicoolmax on 22 Jul 2011, 19:58, edited 2 times in total.



Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2008

Re: Problem solving  3 groups of 3 [#permalink]
Show Tags
22 Jul 2011, 19:02
bschool83 wrote: In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?
280 1,260 1,680 2,520 3,360
I am looking for an explanation, not the answer. \(\frac{(3*3)!}{(3!)^3}=\frac{9!}{6^3}=1680\) Ans: "C"
_________________
~fluke
GMAT Club Premium Membership  big benefits and savings



Manager
Joined: 07 Jun 2011
Posts: 73

Re: Problem solving  3 groups of 3 [#permalink]
Show Tags
22 Jul 2011, 19:19
1
This post received KUDOS
I think the answer is 280 based on the following rationale:
Number of ways for Selecting 3 people out of 9 is 9C3
Number of ways of selecting the next group of 3 is 6C3
Total Number if ways to order three groups is 3! = 6
there for the answer is 9C3*6C3 / 6 = 280
Fluke, have you considered the formula where the order is not important?



Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2008

Re: Problem solving  3 groups of 3 [#permalink]
Show Tags
22 Jul 2011, 19:53
manishgeorge wrote: I think the answer is 280 based on the following rationale:
Number of ways for Selecting 3 people out of 9 is 9C3
Number of ways of selecting the next group of 3 is 6C3
Total Number if ways to order three groups is 3! = 6
there for the answer is 9C3*6C3 / 6 = 280
Fluke, have you considered the formula where the order is not important? Good catch: Should be: 1680/3!=280
_________________
~fluke
GMAT Club Premium Membership  big benefits and savings




Re: Problem solving  3 groups of 3
[#permalink]
22 Jul 2011, 19:53






