It is currently 20 Oct 2017, 07:44

### 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 four distinct pairs of brothers and sisters. In

Author Message
VP
Joined: 22 Nov 2007
Posts: 1079

Kudos [?]: 678 [0], given: 0

There are four distinct pairs of brothers and sisters. In [#permalink]

### Show Tags

27 Jan 2008, 11:25
1
This post was
BOOKMARKED
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

There are four distinct pairs of brothers and sisters. In how many ways can a committee of 3 be formed and NOT have siblings in it?
 8
 24
 32
 56
 80

How can we solve this one directly with combs?

Kudos [?]: 678 [0], given: 0

Director
Joined: 12 Jul 2007
Posts: 857

Kudos [?]: 330 [0], given: 0

### Show Tags

28 Jan 2008, 04:59
Consider each pair a distinct object.
4C3 = 4
Then multiply by 2 for each person chosen since each object has 2 people to choose from.
2*2*2 = 8
4*8 = 32

Kudos [?]: 330 [0], given: 0

Director
Joined: 01 Jan 2008
Posts: 618

Kudos [?]: 198 [0], given: 1

### Show Tags

28 Jan 2008, 07:17
eschn3am wrote:
Consider each pair a distinct object.
4C3 = 4
Then multiply by 2 for each person chosen since each object has 2 people to choose from.
2*2*2 = 8
4*8 = 32

Good solution, eschn3am. Since marcodonzelli wants a combinations formula, here it is: 4C3*(2C1)^3

Kudos [?]: 198 [0], given: 1

CIO
Joined: 02 Oct 2007
Posts: 1218

Kudos [?]: 979 [0], given: 334

### Show Tags

28 Jan 2008, 10:55

7-p395941
_________________

Welcome to GMAT Club!

Want to solve GMAT questions on the go? GMAT Club iPhone app will help.
Result correlation between real GMAT and GMAT Club Tests
Are GMAT Club Test sets ordered in any way?

Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.

GMAT Club Premium Membership - big benefits and savings

Kudos [?]: 979 [0], given: 334

VP
Joined: 22 Nov 2007
Posts: 1079

Kudos [?]: 678 [0], given: 0

### Show Tags

28 Jan 2008, 11:15
dzyubam wrote:

7-p395941

8c3 are the total number of combinations
4c1 represent the choice of 1 couple od siblings
6 are the people who can be chosen to occupy the third place

8c3-4c1*6=56-24=32

Pick one person out of 8, then one out of 6 (the first person with her or his sibling excluded), then one out of 4, giving 8*6*4 permutations. These are called permutations because we count every distinct combination of 3 people 6 times (6=3!). For example, we count {Ann, Bill, and Jane} as {Ann, Bill, Jane}, {Ann, Jane, Bill}, {Jane, Ann, Bill}, {Jane, Bill, Ann}, {Bill, Ann, Jane}, {Bill, Jane, Ann}. To get the number of combinations, we divide the permutations by 6, to get (8*6*4)/6 = 32 combinations

I am not sure about its quickness.

Kudos [?]: 678 [0], given: 0

Senior Manager
Joined: 03 Nov 2005
Posts: 380

Kudos [?]: 68 [0], given: 17

Location: Chicago, IL

### Show Tags

31 Oct 2009, 19:24
My approach is to choose the total possible number of ways and exclude undesired ones.

The number of ways to choose 3 people out of 8 is C(8,3), order does not

and the number of ways to form a group of 3 from a married couple + 6 more people, it can be restated as in how many ways we can select one person out of 6 remaining ones. The total number of ways to select 1 person out of 6 and there are 4 different married couples.

Soln: C(8,3) - 4*C(6,1) =32
_________________

Hard work is the main determinant of success

Kudos [?]: 68 [0], given: 17

Re: combinations and permutations   [#permalink] 31 Oct 2009, 19:24
Display posts from previous: Sort by