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From a group of 3 signers and 3 comedians, a show organizer [#permalink]

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02 Oct 2009, 12:03

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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.
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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.
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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.
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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
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Re: From a group of 3 signers and 3 comedians, a show organizer [#permalink]

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29 Jun 2015, 10:12

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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).

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Re: From a group of 3 signers and 3 comedians, a show organizer [#permalink]

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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
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25 Dec 2015, 13:10

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