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Eight people are to be divided into two groups. What is the [#permalink]
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17 Jul 2007, 09:07
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Eight people are to be divided into two groups. What is the probability that there will be 4 in each groups?



VP
Joined: 08 Jun 2005
Posts: 1145

choices are:
1,7
2,6
3,5
4,4
5,3
6,2
7,1
so the answer should be 1/7
note  8C4/(2^8) is a number greater then 1 !!! since 2,520/256 is greater then 1.



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Joined: 03 Jun 2007
Posts: 376

KillerSquirrel wrote: choices are: 1,7 2,6 3,5 4,45,3 6,2 7,1 so the answer should be 1/7 note  8C4/(2^8) is a number greater then 1 !!! since 2,520/256 is greater then 1.
I missed typing the entire response. It is 8C4/(2^8) * 4C4/(2^8)
Also Killer u missed out the possibilities of 0, 8 and 8,0. Moreover u missed out the possibilites of 8 people being divided into more than 2 groups.



VP
Joined: 08 Jun 2005
Posts: 1145

dahcrap wrote: Also Killer u missed out the possibilities of 0, 8 and 8,0. Moreover u missed out the possibilites of 8 people being divided into more than 2 groups.
1. The stem clearly state " Eight people are to be divided into two groups. What is the probability that there will be 4 in each groups?"
2. For me 0,8 means that you don't have two groups only one group of eight people. I don't like philosophical questions. (i.e. "what is a group of people ?").
3. what's the OA.



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KillerSquirrel wrote: dahcrap wrote: Also Killer u missed out the possibilities of 0, 8 and 8,0. Moreover u missed out the possibilites of 8 people being divided into more than 2 groups.
1. The stem clearly state " Eight people are to be divided into two groups. What is the probability that there will be 4 in each groups?" 2. For me 0,8 means that you don't have two groups only one group of eight people. I don't like philosophical questions. (i.e. "what is a group of people ?"). 3. what's the OA.
Sorry for the confusion. I think I explained it the wrong way.
THere is my actual expanation.
Dividing people into 2 groups of 4 requires choosing 4 people from 8. So 8C4 is the number of the ways of forming 2 groups of 4. So far so good.
Now for the total number of ways of forming 2 groups. This can be as you said groups of 1,7, and 7,1 2,6 and 6,2 3,5 and 5,3 4,4 and the philosophical 0,8 and 8,0. Now I feel we need to calculate the number of ways we can get these combinations. So for 1,7 it is 8C1*8C7. Similarly for 2,6 it is 8C2*8C6, for 5,3 it is 8C3*8C5, for 8,0 it is 8C0*8C8 and for 4,4 it is 8C4*8C4
Now on adding these we get (8C1)^2 + (8C2)^2 + (8C3)^2 + (8C4)^2 and the philosophical (8C0)^2
I guess the answer wud be 8C4/((8C1)^2 + (8C2)^2 + (8C3)^2 + (8C4)^2 and the philosophical (8C0)^2)
Now to take it one step further if the order of the groups matters then it is
8C4/2*((8C1)^2 + (8C2)^2 + (8C3)^2 + (8C4)^2 and the philosophical (8C0)^2)



VP
Joined: 08 Jun 2005
Posts: 1145

dahcrap wrote: KillerSquirrel wrote: dahcrap wrote: Also Killer u missed out the possibilities of 0, 8 and 8,0. Moreover u missed out the possibilites of 8 people being divided into more than 2 groups.
1. The stem clearly state " Eight people are to be divided into two groups. What is the probability that there will be 4 in each groups?" 2. For me 0,8 means that you don't have two groups only one group of eight people. I don't like philosophical questions. (i.e. "what is a group of people ?"). 3. what's the OA. Sorry for the confusion. I think I explained it the wrong way. THere is my actual expanation. Dividing people into 2 groups of 4 requires choosing 4 people from 8. So 8C4 is the number of the ways of forming 2 groups of 4. So far so good. Now for the total number of ways of forming 2 groups. This can be as you said groups of 1,7, and 7,1 2,6 and 6,2 3,5 and 5,3 4,4 and the philosophical 0,8 and 8,0. Now I feel we need to calculate the number of ways we can get these combinations. So for 1,7 it is 8C1*8C7. Similarly for 2,6 it is 8C2*8C6, for 5,3 it is 8C3*8C5, for 8,0 it is 8C0*8C8 and for 4,4 it is 8C4*8C4 Now on adding these we get (8C1)^2 + (8C2)^2 + (8C3)^2 + (8C4)^2 and the philosophical (8C0)^2 I guess the answer wud be 8C4/((8C1)^2 + (8C2)^2 + (8C3)^2 + (8C4)^2 and the philosophical (8C0)^2) Now to take it one step further if the order of the groups matters then it is 8C4/2*((8C1)^2 + (8C2)^2 + (8C3)^2 + (8C4)^2 and the philosophical (8C0)^2)
Hello dahcrap  I like your thinking but I recommend that you will use the "occam's razor" on this one  don't forget you have only two minutes !!!
http://www.answers.com/occam's%20razor



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Posts: 376

KillerSquirrel wrote: dahcrap wrote: KillerSquirrel wrote: dahcrap wrote: Also Killer u missed out the possibilities of 0, 8 and 8,0. Moreover u missed out the possibilites of 8 people being divided into more than 2 groups.
1. The stem clearly state " Eight people are to be divided into two groups. What is the probability that there will be 4 in each groups?" 2. For me 0,8 means that you don't have two groups only one group of eight people. I don't like philosophical questions. (i.e. "what is a group of people ?"). 3. what's the OA. Sorry for the confusion. I think I explained it the wrong way. THere is my actual expanation. Dividing people into 2 groups of 4 requires choosing 4 people from 8. So 8C4 is the number of the ways of forming 2 groups of 4. So far so good. Now for the total number of ways of forming 2 groups. This can be as you said groups of 1,7, and 7,1 2,6 and 6,2 3,5 and 5,3 4,4 and the philosophical 0,8 and 8,0. Now I feel we need to calculate the number of ways we can get these combinations. So for 1,7 it is 8C1*8C7. Similarly for 2,6 it is 8C2*8C6, for 5,3 it is 8C3*8C5, for 8,0 it is 8C0*8C8 and for 4,4 it is 8C4*8C4 Now on adding these we get (8C1)^2 + (8C2)^2 + (8C3)^2 + (8C4)^2 and the philosophical (8C0)^2 I guess the answer wud be 8C4/((8C1)^2 + (8C2)^2 + (8C3)^2 + (8C4)^2 and the philosophical (8C0)^2) Now to take it one step further if the order of the groups matters then it is 8C4/2*((8C1)^2 + (8C2)^2 + (8C3)^2 + (8C4)^2 and the philosophical (8C0)^2) Hello dahcrap  I like your thinking but I recommend that you will use the "occam's razor" on this one  don't forget you have only two minutes !!! http://www.answers.com/occam's%20razor
I hope we dont get these monsters in the actual test. Now Ajay_GMAt can u post the OA and OE.



VP
Joined: 08 Jun 2005
Posts: 1145

dahcrap wrote: I hope we dont get these monsters in the actual test. Now Ajay_GMAt can u post the OA and OE.
I don't think there is an OA or OE for that one. This isn't a GMAT type question. I guess It's from some kind of math text book.



Manager
Joined: 14 Mar 2007
Posts: 60

the OA is 35/81 .. source : PS.zip got from the GMAT club forum



Manager
Joined: 08 Jul 2007
Posts: 173

dahcrap wrote: ajay_gmat wrote: the OA is 35/81 .. source : PS.zip got from the GMAT club forum I am stumped to say the least
There are four different ways:
1, 7
2, 6
3, 5
4, 4
So the answer is 8C4/(8C4+8C3+8C2+8C1)=35/81










