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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

00:00

A

B

C

D

E

Difficulty:

(N/A)

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.
_________________

Right, my bad, but how do I calculate for each group?

6!/(6-4)! is only correct if you can have 3 comedians and 3 singers, right?

Would it be something like:

6!/(6-4)! - 4*[4!/(4-3)!] = 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.
_________________

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.
_________________

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.

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"

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?

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 )

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.

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.

Case 1. If we consider that one performer after another --> 4!*3C2*3C2=216 And I think that this it what was meant in the question.

Case 2. If we rearrange problem and say one group after another, the answer would be: 2!*3C2*3C2=18
_________________

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.
_________________

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

gmatclubot

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

There’s something in Pacific North West that you cannot find anywhere else. The atmosphere and scenic nature are next to none, with mountains on one side and ocean on...

This month I got selected by Stanford GSB to be included in “Best & Brightest, Class of 2017” by Poets & Quants. Besides feeling honored for being part of...

Joe Navarro is an ex FBI agent who was a founding member of the FBI’s Behavioural Analysis Program. He was a body language expert who he used his ability to successfully...