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From a group of 3 signers and 3 comedians, a show organizer

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From a group of 3 signers and 3 comedians, a show organizer [#permalink] New post 02 Oct 2009, 11:03
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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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Re: Combinatorics [#permalink] New post 02 Oct 2009, 12:04
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!/(3-2)! = 6 total permutations per "team" and 36 total permutations.

Anyone feel free to correct me!
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Re: Combinatorics [#permalink] New post 02 Oct 2009, 12:29
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!/(3-2)! = 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!
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Re: Combinatorics [#permalink] New post 02 Oct 2009, 13:07
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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Re: Combinatorics [#permalink] New post 02 Oct 2009, 13:51
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.
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Re: Combinatorics [#permalink] New post 03 Oct 2009, 10:56
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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.
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Re: Combinatorics [#permalink] New post 05 Oct 2009, 20:43
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?
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Re: Combinatorics [#permalink] New post 05 Oct 2009, 20:52
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"
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Re: Combinatorics [#permalink] New post 07 Oct 2009, 10:43
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?
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Re: Combinatorics [#permalink] New post 07 Oct 2009, 11:06
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
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Re: Combinatorics [#permalink] New post 07 Oct 2009, 20:15
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Re: Combinatorics [#permalink] New post 08 Oct 2009, 06:00
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!
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Re: Combinatorics [#permalink] New post 08 Oct 2009, 07:19
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.
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Re: Combinatorics [#permalink] New post 08 Oct 2009, 08:42
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ssandeepan wrote:
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: Combinatorics [#permalink] New post 07 Jan 2013, 10:47
select two singers and two comedians - 3C2* 3C2
arrange them - 4!
Ans - 4!*3*3
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Re: Combinatorics [#permalink] New post 29 Dec 2013, 14:23
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

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Re: Combinatorics   [#permalink] 29 Dec 2013, 14:23
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