It is currently 19 Nov 2017, 05:48

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Eight people are to be divided into two groups. What is the

Author Message
Manager
Joined: 14 Mar 2007
Posts: 59

Kudos [?]: 1 [0], given: 0

Eight people are to be divided into two groups. What is the [#permalink]

### Show Tags

17 Jul 2007, 09:07
00:00

Difficulty:

(N/A)

Question Stats:

100% (00:05) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Eight people are to be divided into two groups. What is the probability that there will be 4 in each groups?

Kudos [?]: 1 [0], given: 0

VP
Joined: 08 Jun 2005
Posts: 1143

Kudos [?]: 254 [0], given: 0

### Show Tags

17 Jul 2007, 11:24
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.

Kudos [?]: 254 [0], given: 0

Senior Manager
Joined: 03 Jun 2007
Posts: 376

Kudos [?]: 16 [0], given: 0

### Show Tags

17 Jul 2007, 11:31
KillerSquirrel wrote:
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.

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.

Kudos [?]: 16 [0], given: 0

VP
Joined: 08 Jun 2005
Posts: 1143

Kudos [?]: 254 [0], given: 0

### Show Tags

17 Jul 2007, 11:39
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.

Kudos [?]: 254 [0], given: 0

Senior Manager
Joined: 03 Jun 2007
Posts: 376

Kudos [?]: 16 [0], given: 0

### Show Tags

17 Jul 2007, 11:57
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)

Kudos [?]: 16 [0], given: 0

VP
Joined: 08 Jun 2005
Posts: 1143

Kudos [?]: 254 [0], given: 0

### Show Tags

17 Jul 2007, 12:04
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 !!!

Kudos [?]: 254 [0], given: 0

Senior Manager
Joined: 03 Jun 2007
Posts: 376

Kudos [?]: 16 [0], given: 0

### Show Tags

17 Jul 2007, 12:09
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 !!!

I hope we dont get these monsters in the actual test. Now Ajay_GMAt can u post the OA and OE.

Kudos [?]: 16 [0], given: 0

VP
Joined: 08 Jun 2005
Posts: 1143

Kudos [?]: 254 [0], given: 0

### Show Tags

17 Jul 2007, 12:13
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.

Kudos [?]: 254 [0], given: 0

Manager
Joined: 14 Mar 2007
Posts: 59

Kudos [?]: 1 [0], given: 0

### Show Tags

18 Jul 2007, 00:42
the OA is 35/81 .. source : PS.zip got from the GMAT club forum

Kudos [?]: 1 [0], given: 0

Manager
Joined: 08 Jul 2007
Posts: 173

Kudos [?]: 11 [0], given: 0

### Show Tags

22 Jul 2007, 07:37
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

Kudos [?]: 11 [0], given: 0

22 Jul 2007, 07:37
Display posts from previous: Sort by