Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 22 May 2015, 14:34

### 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

# There are 5 married couples and a group of three is to be

Author Message
TAGS:
Manager
Joined: 10 Dec 2005
Posts: 97
Followers: 1

Kudos [?]: 18 [0], given: 0

There are 5 married couples and a group of three is to be [#permalink]  20 Dec 2005, 19:48
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

I solved this. But wanna check if you guys have a shorter method.
_________________

JAI HIND!

Senior Manager
Joined: 05 Oct 2005
Posts: 485
Followers: 1

Kudos [?]: 2 [0], given: 0

Re: PS: P&C [#permalink]  20 Dec 2005, 20:07
JAI HIND wrote:
There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

I solved this. But wanna check if you guys have a shorter method.

I'm not too good at perm/comb, but here goes:

10 ways to pick the first person, 8 ways to pick the second (b/c the 1st person's spouse must be excluded), and 6 ways to pick the 3rd (b/c the 2nd person's spouse must be excluded).

Therefore,

(10*8*6)/3! since order does not matter.

Director
Joined: 14 Sep 2005
Posts: 994
Location: South Korea
Followers: 2

Kudos [?]: 45 [0], given: 0

Total number of cases to pick 3 = 10C3 = 120
Number of cases to pick 3 including one couple = 5C1 * 8C1 = 40

120 - 40 = 80

I guess yb's approach is way better.
_________________

Auge um Auge, Zahn um Zahn !

SVP
Joined: 14 Dec 2004
Posts: 1707
Followers: 1

Kudos [?]: 52 [0], given: 0

Re: PS: P&C [#permalink]  21 Dec 2005, 11:21
cool_jonny009 wrote:
yb wrote:
JAI HIND wrote:
There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

I solved this. But wanna check if you guys have a shorter method.

I'm not too good at perm/comb, but here goes:

10 ways to pick the first person, 8 ways to pick the second (b/c the 1st person's spouse must be excluded), and 6 ways to pick the 3rd (b/c the 2nd person's spouse must be excluded).

Therefore,

(10*8*6)/3! since order does not matter.

Can you explain why you divided by 3!

You divide by 3! in order to get rid of same group of 3 fixed people arranged in different order & keep only one. i.e. X Y Z is actually one group but without division by 3!, all possible combinations of X Y Z will be treated as different groups.

In other words, (10*8*6) gives you total number of possible groups but this is permutation but actually we need combination, so divide by 3!
Senior Manager
Joined: 05 Oct 2005
Posts: 485
Followers: 1

Kudos [?]: 2 [0], given: 0

Re: PS: P&C [#permalink]  21 Dec 2005, 11:30
vivek123 wrote:
cool_jonny009 wrote:
yb wrote:
JAI HIND wrote:
There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

I solved this. But wanna check if you guys have a shorter method.

I'm not too good at perm/comb, but here goes:

10 ways to pick the first person, 8 ways to pick the second (b/c the 1st person's spouse must be excluded), and 6 ways to pick the 3rd (b/c the 2nd person's spouse must be excluded).

Therefore,

(10*8*6)/3! since order does not matter.

Can you explain why you divided by 3!

You divide by 3! in order to get rid of same group of 3 fixed people arranged in different order & keep only one. i.e. X Y Z is actually one group but without division by 3!, all possible combinations of X Y Z will be treated as different groups.

In other words, (10*8*6) gives you total number of possible groups but this is permutation but actually we need combination, so divide by 3!

Yes, I agree. To further clarify, note that the order of XYZ does not matter.
XYZ could be :

XYZ
XZY
YXZ
YZX
ZYX
ZXY
or 3! = 3*2 = 6 ways.

Therefore, we must divide by 3! since we don't care about the order.
Re: PS: P&C   [#permalink] 21 Dec 2005, 11:30
Similar topics Replies Last post
Similar
Topics:
3 There are 5 married couples and a group of three is to be 2 29 Jan 2011, 13:19
7 There are 5 married couples and a group of three is to be 8 19 Jul 2008, 16:53
There are 5 married couples and a group of three is to be 18 03 Jun 2007, 07:08
There are 5 married couples and a group of three is to be 1 09 Oct 2006, 10:21
There are 5 married couples and a group of three is to be 7 28 Jan 2006, 23:27
Display posts from previous: Sort by