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How many ways are there to create four teams of three from a

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Director
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How many ways are there to create four teams of three from a [#permalink]

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New post 05 Aug 2008, 22:30
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

How many ways are there to create four teams of three from a group of twelve students?

please explain

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Director
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Re: combination from veritas prep [#permalink]

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New post 05 Aug 2008, 22:55
12c3*9c3*6c3*3c3/4!

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Re: combination from veritas prep [#permalink]

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New post 05 Aug 2008, 23:01
Agree with Abhijit

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Re: combination from veritas prep [#permalink]

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New post 05 Aug 2008, 23:04
can you supply a brief explanation?

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New post 05 Aug 2008, 23:09
never mind..i get it, but i'm looking for a more intuitive way to do combo problems.

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Re: combination from veritas prep [#permalink]

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New post 06 Aug 2008, 05:09
abhijit_sen wrote:
12c3*9c3*6c3*3c3/4!


could you explain the division by 4!

Thanks..

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Re: combination from veritas prep [#permalink]

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New post 06 Aug 2008, 07:07
Once you get the 4 teams created with 3 members on each team, you do not want to count the order of the teams as different.

Name the teams: Team A, Team B, Team C, Team D.

A-B-C-D is the same as B-A-C-D

Because there are 4 teams, divide by 4! (24)
bhushangiri wrote:
abhijit_sen wrote:
12c3*9c3*6c3*3c3/4!


could you explain the division by 4!

Thanks..

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Re: combination from veritas prep [#permalink]

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New post 06 Aug 2008, 07:19
jallenmorris wrote:
Once you get the 4 teams created with 3 members on each team, you do not want to count the order of the teams as different.

Name the teams: Team A, Team B, Team C, Team D.

A-B-C-D is the same as B-A-C-D

Because there are 4 teams, divide by 4! (24)


But what i am still not clear about is, how does the expression 12C3*9C3*6C3*3C3 count the order..

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Re: combination from veritas prep [#permalink]

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New post 06 Aug 2008, 07:22
think .. i got it..

12C3*9C3*.... is akin to doing n*(n-1)*(n-2) which takes into account an order.... and in these cases you divide by appropriate x! when you want to disregard the order or when there are x of same type. right ?

Thanks

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Re: combination from veritas prep [#permalink]

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New post 06 Aug 2008, 07:29
I'm still slightly confused. What does the c stand for?

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Re: combination from veritas prep [#permalink]

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New post 06 Aug 2008, 07:33
rahulm wrote:
I'm still slightly confused. What does the c stand for?


combination. you will see sometimes see p which stands for permutation.

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Re: combination from veritas prep [#permalink]

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New post 06 Aug 2008, 08:29
No. of ways to form 1st team: 12*11*10= 1320
No. of ways to form 2nd team: 9*8*7=504
No. of ways to form 3rd team: 6*5*4=120
No. of ways to form 4th team: 3*2*1=6

therefore total: 1950 (ans)

Whats the OA/OE?

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Re: combination from veritas prep [#permalink]

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New post 06 Aug 2008, 08:56
dishant007 wrote:
No. of ways to form 1st team: 12*11*10= 1320
No. of ways to form 2nd team: 9*8*7=504
No. of ways to form 3rd team: 6*5*4=120
No. of ways to form 4th team: 3*2*1=6

therefore total: 1950 (ans)

Whats the OA/OE?


this is incorrect you are taking permutaitons. Here order is not important.
No. of ways to form 1st team: 12*11*10= 1320
it should be 12c3

take another example if there 3 persons and need to form team of 3 person. how many ways you can do it.
PEROSNS A B C
only one team can be formed .. ABC
here order doesn't matter ABC=BCA=CBA=..
3C3

If question asks how do you arange 3 persons in 3 seats.
then 3!
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Re: combination from veritas prep   [#permalink] 06 Aug 2008, 08:56
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