Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 03 Nov 2005
Posts: 385
Location: Chicago, IL

From a group of 3 signers and 3 comedians, a show organizer [#permalink]
Show Tags
02 Oct 2009, 12:03
1
This post received KUDOS
2
This post was BOOKMARKED
Question Stats:
38% (01:26) correct
63% (01:02) wrong based on 30 sessions
HideShow timer Statistics
From a group of 3 signers and 3 comedians, a show organizer might select two singers and two comedians to appear one after another in the show, how many different ways an organizer can arrange the performers in the show.
_________________
Hard work is the main determinant of success



Manager
Joined: 30 Sep 2009
Posts: 145

Re: Combinatorics [#permalink]
Show Tags
02 Oct 2009, 13:04
If the question just wants to know how many combinations (regardless of order): 3C2 = 3 combinations for each side and 9 total four person groups for the stage (3*3). If it wants to know the total permutations: 3!/(32)! = 6 total permutations per "team" and 36 total permutations. Anyone feel free to correct me!
_________________
School: Boston College MBA '12



Senior Manager
Joined: 03 Nov 2005
Posts: 385
Location: Chicago, IL

Re: Combinatorics [#permalink]
Show Tags
02 Oct 2009, 13:29
hypermeganet wrote: If the question just wants to know how many combinations (regardless of order):
3C2 = 3 combinations for each side and 9 total four person groups for the stage (3*3).
If it wants to know the total permutations:
3!/(32)! = 6 total permutations per "team" and 36 total permutations.
Anyone feel free to correct me! Incorrect, there are 4 players that can make an appearance on the stage in 4! different orders, not 3!
_________________
Hard work is the main determinant of success



Manager
Joined: 30 Sep 2009
Posts: 145

Re: Combinatorics [#permalink]
Show Tags
02 Oct 2009, 14:07
Right, my bad, but how do I calculate for each group? 6!/(64)! is only correct if you can have 3 comedians and 3 singers, right? Would it be something like: 6!/(64)!  4*[4!/(43)!] = 264 permutations that involve only 2 from each? I could be way off. I am trying to take the total permutations (360) and take out the permutations involving 3 from one group and only one from the other.
_________________
School: Boston College MBA '12



Senior Manager
Joined: 21 Jul 2009
Posts: 364
Schools: LBS, INSEAD, IMD, ISB  Anything with just 1 yr program.

Re: Combinatorics [#permalink]
Show Tags
02 Oct 2009, 14:51
rlevochkin wrote: From a group of 3 signers and 3 comedians, a show organizer might select two singers and two comedians to appear one after another in the show, how many different ways an organizer can arrange the performers in the show.
Use different methods and explain them The question stem clearly states that the organizer is "selecting" two singers and two comedians to appear one after another. Total 4 are selected from group of 6 so 6C4 = 15 possible combinations of 4 entertainers together. What if "one after another" actually meant, a singer and comedian alternatively and 4 such entertainers for the evening? If that is the case, it would be 2 X 3C2 X 3C2 = 18 possible combinations, which apparently gives the organizer more work in choosing the entertainers.
_________________
I am AWESOME and it's gonna be LEGENDARY!!!



Manager
Joined: 25 Aug 2009
Posts: 168
Location: Streamwood IL
Schools: Kellogg(Evening),Booth (Evening)
WE 1: 5 Years

Re: Combinatorics [#permalink]
Show Tags
03 Oct 2009, 11:56
5
This post received KUDOS
Ways to select 2 Singers out of 3 = 3C2 = 3 Ways to select 2 Comedians out of 3 = 3C2 = 3 Ways to rearrange 4 performers = !4 = 24 Total combinations = 3*3*24=216.
_________________
Rock On



Intern
Joined: 05 Oct 2009
Posts: 32
Location: Hamamatsu, Japan
Schools: Attending IE Nov 2010
WE 1: 11 years int'l work experience

Re: Combinatorics [#permalink]
Show Tags
05 Oct 2009, 21:43
rlevochkin wrote: From a group of 3 signers and 3 comedians, a show organizer might select two singers and two comedians to appear one after another in the show, how many different ways an organizer can arrange the performers in the show.
Use different methods and explain them I agree with Atish's formula, and that was my first answer as well, but I'm just thinking... If we visualize the possible slots as: ABC DEF _ _ _ _ Then in theory we could say the first choice has 6 options and the second choice would be 5 options. If the manager chose the best comedian and the best singer first, the manager should then have 4 possible choices for the 3rd slot. However, if the manager chose two comedians first, then he would only have 3 possible options for the 3rd slot. Either way he would have 2 options for the final pick. If we wanted to maximize the possible choices, we would have to assume he picked one from each in order. That however, would lead us to 6*5*4*2 = 240. Am I wrong?



Intern
Joined: 05 Oct 2009
Posts: 32
Location: Hamamatsu, Japan
Schools: Attending IE Nov 2010
WE 1: 11 years int'l work experience

Re: Combinatorics [#permalink]
Show Tags
05 Oct 2009, 21:52
Also, considering the original question didn't stipulate picking order and asked for multiple methods, I would assume that the largest accurate number of possibilities would be the more precise answer since we're looking for a "maximum possible"



Current Student
Joined: 12 Nov 2008
Posts: 367
Schools: Ross (R2), Cornell (R3) , UNC (R3) , INSEAD (R1 Jan)
WE 1: Advisory (2 yrs)
WE 2: FP & Analysis (2 yrs at matriculation)

Re: Combinatorics [#permalink]
Show Tags
07 Oct 2009, 11:43
The way I'm reading this (correct me if I'm wrong) is that the order of the performers matters (he is trying to create a show), so this is a permutation. The solutions presented so far assume that order doesn't matter. Thoughts?



Manager
Joined: 09 Jul 2007
Posts: 243

Re: Combinatorics [#permalink]
Show Tags
07 Oct 2009, 12:06
I think it should be 72.
Reason :
1 )The organizer can select the 1st performer out of any of the 6 performers. 2 )Now, the 2nd position can be filled by 3 ways ( from the opposite group of the 1st performer. eg If the 1st perfomer is a comedians, the 2nd performer will be any of the 3 singers ) 3 ) 3rd position can be filled in by 2 ways ( becase 1 performer is already selected as 1st performer. 4 ) 4th position can be also filled by 2 ways ( 1 performer already seletcted as 2nd performer )
So, total ways = 6 *3*2*2 = 72



Founder
Joined: 04 Dec 2002
Posts: 15256
Location: United States (WA)

Re: Combinatorics [#permalink]
Show Tags
07 Oct 2009, 21:15
_________________
Founder of GMAT Club
US News Rankings progression  last 10 years in a snapshot  New! Just starting out with GMAT? Start here... Need GMAT Book Recommendations? Best GMAT Books
Coauthor of the GMAT Club tests



Manager
Joined: 18 Jul 2009
Posts: 51

Re: Combinatorics [#permalink]
Show Tags
08 Oct 2009, 07:00
Its not clear from the question that two singer and two comedian gonna perform as a individual or in a group of two.
if 4 are going perform separately then atish is right.
ans will be 3c2*3c2*4!



Manager
Joined: 09 Jul 2007
Posts: 243

Re: Combinatorics [#permalink]
Show Tags
08 Oct 2009, 08:19
one after another  is little confisuing to me. I thought it meant commedian after singer , or singer after commedian. But looks like it is trying to mean one performer after another, in whcih case 3C2*3C2*4 ! should be the answer.



Math Expert
Joined: 02 Sep 2009
Posts: 40403

Re: Combinatorics [#permalink]
Show Tags
08 Oct 2009, 09:42



Manager
Joined: 13 Oct 2012
Posts: 70
Concentration: General Management, Leadership

Re: Combinatorics [#permalink]
Show Tags
07 Jan 2013, 11:47
select two singers and two comedians  3C2* 3C2 arrange them  4! Ans  4!*3*3



Current Student
Joined: 06 Sep 2013
Posts: 1997
Concentration: Finance

Re: Combinatorics [#permalink]
Show Tags
29 Dec 2013, 15:23
atish wrote: Ways to select 2 Singers out of 3 = 3C2 = 3 Ways to select 2 Comedians out of 3 = 3C2 = 3 Ways to rearrange 4 performers = !4 = 24
Total combinations = 3*3*24=216. Agree with you my friend, did it the exact same way Kudos for you! Cheers! J



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16573

Re: From a group of 3 signers and 3 comedians, a show organizer [#permalink]
Show Tags
29 Jun 2015, 10:12
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



Intern
Joined: 04 Sep 2015
Posts: 36
Location: Germany
Concentration: Operations, Finance
WE: Project Management (Aerospace and Defense)

Re: From a group of 3 signers and 3 comedians, a show organizer [#permalink]
Show Tags
25 Dec 2015, 13:10
May be, this post is too old..
different approach  we can also FCP (fundamental counting principle)
There are 4 slots (with restriction of one singer after comedian)
Place 1 (choose any)  lets choose comedian  3 ways (3 comedians, and anyone can play) Place 2  it has to be singer (from question)  again, 3 ways (3 singers can be chosen in 3 ways) place 3  comedian  2 ways (out of 2 comedians left) Place 4  singer  2 ways (out of 2 singers left)
So, total ways = 3 x 3 x 2 x 2 = 36 ways
However, order in which singer & comedian can be calculated using MISSISSIPPI rule = 4! / (2! x 2!) = 6 ways
Total ways = 36 x 6 = 218 ways




Re: From a group of 3 signers and 3 comedians, a show organizer
[#permalink]
25 Dec 2015, 13:10







