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In how many ways can 12 students be partitioned into 3

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In how many ways can 12 students be partitioned into 3 [#permalink] New post 20 Aug 2004, 07:34
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In how many ways can 12 students be partitioned into 3 teams, A, B, and C, so that each team contains 4 students ?
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Ps -ways [#permalink] New post 20 Aug 2004, 08:51
In how many ways can 12 students be partitioned into 3 teams, A, B, and C, so that each team contains 4 students ?

Formula: If 2m things are to be divided into 2 groups,each containing mthings , the number of ways = (2m)!/2!(m!)^2

hence here m = 4

(3*m)!/3!(m!)^3
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 [#permalink] New post 20 Aug 2004, 12:05
smandalika wrote:
12C4* 8C4
:no
give it another shot.
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 [#permalink] New post 20 Aug 2004, 12:07
I posted the previous message. Can anyone else explain this problem ?
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 [#permalink] New post 20 Aug 2004, 22:55
12C4 x 8C4 / 3! = 5775?

12C4 = to select first 4 out of 12 for the first group
8C4 = to select second 4 out of the remaining 8 for the second group
the remaining guys will form a group within themselves.

these three groups are interchangeable. So divide by 3! ways.
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 [#permalink] New post 21 Aug 2004, 17:14
hardworker_indian wrote:
12C4 x 8C4 / 3! = 5775?

these three groups are interchangeable. So divide by 3! ways.


I'm not sure what u mean by this , coz I'm thinking we are just talking combinations and not arrangement. Could you explain in detail.
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 [#permalink] New post 21 Aug 2004, 17:36
I got 12C4*8C4*3!
I included times 3! because a particular set of students could fall into group A, group B, or group C
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 [#permalink] New post 21 Aug 2004, 18:15
12C4*8C4/3! should be right. This is because the first group of 4 that you select could have been part of group A, B or C. Hence, any 4 persons selected could have been "interchanged" into any of A, B or C.
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 [#permalink] New post 21 Aug 2004, 20:38
(12C4 X 8C4 )/4 = 5775 is correct.

But, I am still not able to understand this to the extent of applying the same logic for similar problems.

Are you saying - members from one group shouldn't appear in other groups.
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 [#permalink] New post 21 Aug 2004, 20:39
smandalika wrote:
hardworker_indian wrote:
12C4 x 8C4 / 3! = 5775?

these three groups are interchangeable. So divide by 3! ways.


I'm not sure what u mean by this , coz I'm thinking we are just talking combinations and not arrangement. Could you explain in detail.


smandalika,
Not sure if I can explain clearly, but will try. My answer is:
12C4 x 8C4 - gives you the no of ways we can arrange 12 guys in group A (say Maths group, 4 guys), group B (say science group, 4 guys) and group C (say english group, 4 guys)

12C4 x 8C4 x 3! - gives the number of ways these three groups (Maths, Scinece and English) can be arranged (say, placing these three groups in three cars - green car, blue car and red car)

Actual problem does not differentiate between the three groups. All these are meant for the same purpose and are interchangeable. No group is assigned any specific role or character.
Hence, we have to divide by the number of groups!, since these groups are interchangeable (I visualize these three groups to be just standing in chunks on the same platform).

My notes:
=> No of ways of giving r1 things to one group and r2 things to one group from n objects (n=r1 + r2) = nCr1 = n!/(r1!*r2!)
(When u select r1 out of the group r2 is left behind by itself.)
=> for three groups = (m+n+p)!/(m! x n! x p!)
=> for three equal groiups = (3m)!/(m!)^3
=> for three intechangeable groups = (3m!)/(m!^3 x 3!)

I am sure this wasn't cliear enough, but wanted to write my thoughts; I will request Akamai and Ian to give good theory.
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 [#permalink] New post 21 Aug 2004, 23:36
The formula for 3 groups of 4 out of 12 total is simply 12!/(4!4!4!).

There are a lot of ways to derive this but here is one that is a little long, but is easy to understand intuitively.

Think of the problem as 3 sequential tasks. First, we need to get the first group of 4 or 12C4. Next we need to find the next group of 4 from the remainig 3 so that is 8C4. Finally, there is 4C4 ways to choose 4 from 4. Hence, we have:

12!/(4!8!) * 8!/4!4! * 4!/4!0! = 12!/(4!4!4!).

Or you can think of it this way. There are 12 ways to pick the first person, then 11, 10 and 9 ways to pick the next 3. But we don't care about the order so we divide by 4! (i.e., there are 4! ways a specific combination of 4 unique numbers can be combined into ordered sequences, but we don't care about order so we divide out this number). Hence we get 12*11*10*9/4!. There are 8, 7, 6, and 5 ways to choose the next 4 people, and once again we divide by 4! so we have 8*7*6*5/4!. Finally, we have 4*3*2*1 ways choose the next 4, but we once again divide by 4! (makes sense as there is only ONE way you can choose a combination of 4 people from 4 people). Multiply this together, you get 12*11*10*9/4! * 8*7*6*5/4! * 4*3*2*1/4! or 12!/(4!4!4!).

Hope this helps,

AB
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 [#permalink] New post 22 Aug 2004, 08:11
AkamaiBrah wrote:
The formula for 3 groups of 4 out of 12 total is simply 12!/(4!4!4!).

There are a lot of ways to derive this but here is one that is a little long, but is easy to understand intuitively.

Think of the problem as 3 sequential tasks. First, we need to get the first group of 4 or 12C4. Next we need to find the next group of 4 from the remainig 3 so that is 8C4. Finally, there is 4C4 ways to choose 4 from 4. Hence, we have:

12!/(4!8!) * 8!/4!4! * 4!/4!0! = 12!/(4!4!4!).

Or you can think of it this way. There are 12 ways to pick the first person, then 11, 10 and 9 ways to pick the next 3. But we don't care about the order so we divide by 4! (i.e., there are 4! ways a specific combination of 4 unique numbers can be combined into ordered sequences, but we don't care about order so we divide out this number). Hence we get 12*11*10*9/4!. There are 8, 7, 6, and 5 ways to choose the next 4 people, and once again we divide by 4! so we have 8*7*6*5/4!. Finally, we have 4*3*2*1 ways choose the next 4, but we once again divide by 4! (makes sense as there is only ONE way you can choose a combination of 4 people from 4 people). Multiply this together, you get 12*11*10*9/4! * 8*7*6*5/4! * 4*3*2*1/4! or 12!/(4!4!4!).

Hope this helps,

AB


You have missed one 3! in this.

12!/(4!4!4!3!) -is the answer. I am not clear on that 3!.
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 [#permalink] New post 22 Aug 2004, 09:56
The general formula for the number of ways in which (m+n+p) things can be divided in to 3 groups containing m,n,p things =
(m+n+p)! / (m! * n! * p!)
In this case, m = n =p (as the teams have equal number of students i.e 4)
Hence, the Total number of ways (considering the order also) =
(4+4+4)! / [4! * 4! * 4!]

Since there are 3! such orders (as the number of teams is 3).
As all these orders should not be counted as different (according to problem definition), we are required to divide by 3!.

Hence the answer = 12! / (4! 4! 4! 3!) = 5775 ways
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 [#permalink] New post 23 Aug 2004, 00:03
wireless_neo wrote:
The general formula for the number of ways in which (m+n+p) things can be divided in to 3 groups containing m,n,p things =
(m+n+p)! / (m! * n! * p!)
In this case, m = n =p (as the teams have equal number of students i.e 4)
Hence, the Total number of ways (considering the order also) =
(4+4+4)! / [4! * 4! * 4!]

Since there are 3! such orders (as the number of teams is 3).
As all these orders should not be counted as different (according to problem definition), we are required to divide by 3!.

Hence the answer = 12! / (4! 4! 4! 3!) = 5775 ways


NO!

The three teams are designated A, B, and C.
Hence, you do NOT divide by 3!
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 [#permalink] New post 23 Aug 2004, 04:10
Welcome back AkamaiBrah :)

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 [#permalink] New post 23 Aug 2004, 05:30
Akamai, When would the groups be considered as interchangeable? (like dividing 12 apples equally among three baskets - should the 3! apply in this case).
Thanks.
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 [#permalink] New post 23 Aug 2004, 07:26
AkamaiBrah wrote:
wireless_neo wrote:
The general formula for the number of ways in which (m+n+p) things can be divided in to 3 groups containing m,n,p things =
(m+n+p)! / (m! * n! * p!)
In this case, m = n =p (as the teams have equal number of students i.e 4)
Hence, the Total number of ways (considering the order also) =
(4+4+4)! / [4! * 4! * 4!]

Since there are 3! such orders (as the number of teams is 3).
As all these orders should not be counted as different (according to problem definition), we are required to divide by 3!.

Hence the answer = 12! / (4! 4! 4! 3!) = 5775 ways


NO!

The three teams are designated A, B, and C.
Hence, you do NOT divide by 3!


I agree completely. For anyone who's read any of my posts, you know I don't like using the combinations jargon (it doesn't help anyone who reads this and doesn't know what it's about!). But I do see this as a combinations question within the teams, just not across all three teams. So my paper looks like this:

Image

AB already explained this perfectly in the second part of his first posting, so I won't go into detail about it. If anyone has questions, let me know.

Hope that helps!
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 [#permalink] New post 23 Aug 2004, 17:51
AkamaiBrah wrote:
wireless_neo wrote:
The general formula for the number of ways in which (m+n+p) things can be divided in to 3 groups containing m,n,p things =
(m+n+p)! / (m! * n! * p!)
In this case, m = n =p (as the teams have equal number of students i.e 4)
Hence, the Total number of ways (considering the order also) =
(4+4+4)! / [4! * 4! * 4!]

Since there are 3! such orders (as the number of teams is 3).
As all these orders should not be counted as different (according to problem definition), we are required to divide by 3!.

Hence the answer = 12! / (4! 4! 4! 3!) = 5775 ways


NO!

The three teams are designated A, B, and C.
Hence, you do NOT divide by 3!


Oops, I didn't see the A, B and C from the question when I answered :cry:
Thanks Akamai and Ian.

Geethu, do you have the source of this problem? And any OE?
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 [#permalink] New post 23 Aug 2004, 17:56
Quote:
Oops, I didn't see the A, B and C from the question when I answered :cry:
Thanks Akamai and Ian.

Geethu, do you have the source of this problem? And any OE?


That was me, logged in now.
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 [#permalink] New post 24 Aug 2004, 05:41
smandalika wrote:
12C4* 8C4


This posting from above is correct, by the way.
  [#permalink] 24 Aug 2004, 05:41
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