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How many ways are there to split a group of 6 boys into two

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How many ways are there to split a group of 6 boys into two [#permalink] New post 26 Aug 2008, 09:20
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How many ways are there to split a group of 6 boys into two groups of 3 boys each? (The order of the groups does not matter)


* 8
* 10
* 16
* 20
* 24

why is the ans not 6c3?

suppose the quesiton was 3 groups of 2 boys each, what would be the ans?
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Re: gmat club [#permalink] New post 26 Aug 2008, 09:31
arjtryarjtry wrote:
How many ways are there to split a group of 6 boys into two groups of 3 boys each? (The order of the groups does not matter)


* 8
* 10
* 16
* 20
* 24

why is the ans not 6c3?

suppose the quesiton was 3 groups of 2 boys each, what would be the ans?


Answer is 6C3 = 20

For 3 groups of 2 boys it will be 6C2*4C2*2C2 = 15*6 = 90
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Re: gmat club [#permalink] New post 26 Aug 2008, 09:32
arjtryarjtry wrote:
How many ways are there to split a group of 6 boys into two groups of 3 boys each? (The order of the groups does not matter)


* 8
* 10
* 16
* 20
* 24

why is the ans not 6c3?

suppose the quesiton was 3 groups of 2 boys each, what would be the ans?


The answer is 6C3/2! = 10

6C3 is right if you are asked to form one group of 3 people.
The actual expression is 6C3*3C3, where 3C3 =1.
But here you are bringing in order - group formed 1st , group formed 2nd. Since order does not matter, divide by the number of wasy the group can be arranged.

If you had to form 3 groups of two ppl each, then

first group can be formed in 6C2 ways. The second group can be formed in 4C2 ways and the third one can be formed in 2C2 ways.
so you have 6C2*4C2*2C2. But this orders the the three groups into 3! ways.
Your answer would be 6C2*4C2*2C2/3! = 15

The "C"s in the 6C2*4C2*2C2 eliminate the ordering of players only. Not the ordering of teams. You need to account for the ordering of teams also to find the true answer.

Hope that helps..

Last edited by bhushangiri on 26 Aug 2008, 09:41, edited 1 time in total.
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Re: gmat club [#permalink] New post 26 Aug 2008, 09:40
bhushangiri wrote:
arjtryarjtry wrote:
How many ways are there to split a group of 6 boys into two groups of 3 boys each? (The order of the groups does not matter)


* 8
* 10
* 16
* 20
* 24

why is the ans not 6c3?

suppose the quesiton was 3 groups of 2 boys each, what would be the ans?


The answer is 6C3/2! = 10

6C3 is right if you are asked to form one group of 3 people.
The is because the actual expression is 6C3*3C3.
But here you are bringing in order - group formed 1st , group formed 2nd. Since order does not matter, divide by the number of wasy the group can be arranged.

If you had to form 3 groups of two ppl each, then

first group can be formed in 6C2 ways. The second group can be formed in 4C2 ways and the third one can be formed in 2C2 ways.
so you have 6C2*4C2*2C2. But this orders the the three groups into 3! ways.
Your answer would be 6C2*4C2*2C2/3! = 15

The "C"s in the 6C2*4C2*2C2 choose only the players. But not the teams. You need to account for the ordering of teams.

Hope that helps..


Right ... keep slipping out of my mind. Last time I calculated rightly , this time wrong
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Re: gmat club [#permalink] New post 26 Aug 2008, 09:52
thanks bhushan.. it helped me.
so we have to divide b the no. of grps thus formed, and multiplication should be done for the same no. of grps is it ?
eg
three grps -> 6c2*4c2*2c2/3!
two grps-> 6c3*3c3/3!

suppose it was 1 group of 4 people
is it just 6c4?
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Re: gmat club [#permalink] New post 26 Aug 2008, 09:53
Also, look from the tallest boy's perspective: 2 of the 5 other boys will be on his team: 5C2 = 10
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Re: gmat club [#permalink] New post 26 Aug 2008, 10:19
arjtryarjtry wrote:
thanks bhushan.. it helped me.
so we have to divide b the no. of grps thus formed, and multiplication should be done for the same no. of grps is it ?
eg
three grps -> 6c2*4c2*2c2/3!
two grps-> 6c3*3c3/3!

suppose it was 1 group of 4 people
is it just 6c4?


Yes, if it is C_6^4 for selecting 1 group of 4 people from 6. The thing that I forgot with regard to this question was the significance of "order does not matter." I was thinking that C_6^3 was correct, but didn't remember to divide by the permutations of the order of the group. For instance.

3 groups of 3 from 9 people

\frac{C_3^9 * C_3^6 * C_3^3}{3!}

The first group formed can have any 3 of the 9. next group will be any 3 of the 6 remaining, and the last group is formed by selection the 3 from the 6, because you have 3 remaining, which is why it will always be 1. But then you have 3 groups, so you need to disregard their order, which means divide by the number of ways these 3 groups can be ordered. Which is 3!

Example:

A B C D E F G H I

CDE | BGI | AFH

This is 1 option. because order does not matter, we need to eliminate situations such as:

BGI | CDE | AFH

This is the same as counting the number of ways 3 things can be arranged. Permutation of 3 things = 3!.

I believe the answer for 3 groups of 3 out of 9 people, and order does not matter would be 280.

C_9^3 = \frac{9*8*7}{3*2*1} * C_6^3 = \frac{6*5*4}{3*2*1} * C_3^3 = 1

This all comes to be
\frac{9*8*7*6*5*4*1}{3*2*1*3*2*1*3*2*1}

8*7*5

7*40

280
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Re: gmat club [#permalink] New post 26 Aug 2008, 11:08
jallenmorris wrote:
arjtryarjtry wrote:
thanks bhushan.. it helped me.
so we have to divide b the no. of grps thus formed, and multiplication should be done for the same no. of grps is it ?
eg
three grps -> 6c2*4c2*2c2/3!
two grps-> 6c3*3c3/3!

suppose it was 1 group of 4 people
is it just 6c4?


Yes, if it is C_6^4 for selecting 1 group of 4 people from 6. The thing that I forgot with regard to this question was the significance of "order does not matter." I was thinking that C_6^3 was correct, but didn't remember to divide by the permutations of the order of the group. For instance.

3 groups of 3 from 9 people

\frac{C_3^9 * C_3^6 * C_3^3}{3!}

The first group formed can have any 3 of the 9. next group will be any 3 of the 6 remaining, and the last group is formed by selection the 3 from the 6, because you have 3 remaining, which is why it will always be 1. But then you have 3 groups, so you need to disregard their order, which means divide by the number of ways these 3 groups can be ordered. Which is 3!

Example:

A B C D E F G H I

CDE | BGI | AFH

This is 1 option. because order does not matter, we need to eliminate situations such as:

BGI | CDE | AFH

This is the same as counting the number of ways 3 things can be arranged. Permutation of 3 things = 3!.

I believe the answer for 3 groups of 3 out of 9 people, and order does not matter would be 280.

C_9^3 = \frac{9*8*7}{3*2*1} * C_6^3 = \frac{6*5*4}{3*2*1} * C_3^3 = 1

This all comes to be
\frac{9*8*7*6*5*4*1}{3*2*1*3*2*1*3*2*1}

8*7*5

7*40

280



good job Allen and Bhusan.


What if question says
How many ways are there to split a group of 6 boys into two groups of 4 boys in one group and 2 boys in another group? (The order of the groups does not matter)
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Re: gmat club [#permalink] New post 26 Aug 2008, 12:00
x2suresh wrote:

good job Allen and Bhusan.


What if question says
How many ways are there to split a group of 6 boys into two groups of 4 boys in one group and 2 boys in another group? (The order of the groups does not matter)


Actually I should not have used the word "order" in the earlier posts. Multiple counting would have been a better choice of word. Double counting when there are two groups, 3! counting when there are 3 groups..... so on and so forth.

Now coming to your question.. two groups - one with 4 ppl and the other with 2 ppl.
In this case the answer will be 6C4*2C2. Here there is no scope of double counting of groups as the two groups are of unequal number. I am not sure if this is clear. But I cant think of any other way except using exaples of the two cases... which is a last resort.
:|
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Re: gmat club [#permalink] New post 26 Aug 2008, 12:04
bhushangiri wrote:
x2suresh wrote:

good job Allen and Bhusan.


What if question says
How many ways are there to split a group of 6 boys into two groups of 4 boys in one group and 2 boys in another group? (The order of the groups does not matter)


Actually I should not have used the word "order" in the earlier posts. Multiple counting would have been a better choice of word. Double counting when there are two groups, 3! counting when there are 3 groups..... so on and so forth.

Now coming to your question.. two groups - one with 4 ppl and the other with 2 ppl.
In this case the answer will be 6C4*2C2. Here there is no scope of double counting of groups as the two groups are of unequal number. I am not sure if this is clear. But I cant think of any other way except using exaples of the two cases... which is a last resort.
:|


agreed. Only when we divide the group into equal number groups....
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Re: gmat club   [#permalink] 26 Aug 2008, 12:04
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