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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

8C4 includes all the combinations = 70
4x6C2 includes all the combinations which is turn includes all the two pairs twice i.e. A1A2B1B2 and B1B2A1A2
so only need to subtract once the 4C2

In this scenario brute force seems easier. I don't think i would have figured out to subtract only once the 4C2 it in the GMAT under pressure

Re: combinatorics - siblings [#permalink]
04 Dec 2007, 06:26

1

This post received KUDOS

bmwhype2 wrote:

There are four distinct pairs of brothers and sisters. In how many ways can a committee of 4 be formed and NOT have siblings in it?

8*6*4*2 / 4! = 384/24 = 16

Just brute force it, on first place you can put 8, on second you can put 6 (excluding 1 sibling) on third you can put 4 (exclude) 2 siblings... then divide by the number of permutations as position doesnt matter.

Re: combinatorics - siblings [#permalink]
04 Dec 2007, 07:53

antihero wrote:

bmwhype2 wrote:

There are four distinct pairs of brothers and sisters. In how many ways can a committee of 4 be formed and NOT have siblings in it?

8*6*4*2 / 4! = 384/24 = 16

Just brute force it, on first place you can put 8, on second you can put 6 (excluding 1 sibling) on third you can put 4 (exclude) 2 siblings... then divide by the number of permutations as position doesnt matter.

i'm not even sure of the answer. maybe walker can help out.

4 pairs = 4*2 = 8 people total

8C4 = 8!/4!4! = 70 total outcomes

Total – unfavorable = favorable

Unfavorable outcomes
Assuming one pair of twins in the committee, we have two spaces left. Since we plugged a pair of twins in the committee, we have 8-2= 6 people to fill 2 spaces.

6C2 = 6!/2!4! = 15 ways to fill the two remaining slots

We only filled the slots with one pair, and we have to account for arrangements of the pairs. Now, we have 4 pairs. 4*15= 60 total arrangements

When we place members into the remaining slots, there may be an additional set of twins. There are 2 remaining slots to which we can fit a pair of twins. If it were one remaining slot, we cannot fit a pair of twins, so we wouldn’t have to account for duplicates.

Now we account for the number of duplicates.
# of duplicates = Total arrangements - # of unique combinations

To find the # of duplicates of twins, we need treat a pair of twins as one unit.
This means the 4 slots are really 2 slots.

Total ways of arranging four pairs of twins in two slots
4P2 = 4*3 = 12 total ways

Total # of unique combinations
Choosing two pairs out of 4 pairs
= 4C2
= 6
Therefore, # of duplicates = 12 - 6 = 6 duplicates

Re: combinatorics - siblings [#permalink]
11 Mar 2009, 22:28

2

This post received KUDOS

another method

The commitee of 4 , NOT having siblings can be formed in following ways:

0 Sisters and 4 brothers = 4C4 =1 + 1 sister and 3 brothers = 4C1*3C3 = 4 [ 3C3 because selected sister's brother can not be among 3 bros] + 2 sisters and 2 brothers = 4C2* 2C2 = 6 [ again 2C2 because brothers of 2 selected sisters can not be on commitee] + 3 sisters and 1 brother = 4C3* 1C1 = 4 [ only one brother whose sister is not on commitee can be selected] + 4 sisters and 0 brothers = 4C4 = 1

To make a committee consist of 4 member out of 4 pair of siblings .lets say A1A2 B1B2 D1D2 E1E2 we have to select one out of each sibling pair no of selection for each pair =2c1=2 we need to select this way from each pair=2C1 (to select from A1A2) * 2C1 (to select from B1B2)*2C1 (D1D2) *2C1(E1E2) =2C1^4=2^4=16

Re: combinatorics - siblings [#permalink]
13 Dec 2009, 02:41

1

This post received KUDOS

8 * 6 * 4 *2 / 4! = 16

Note that 8 * 6 * 4 *2 create a duplicate such as ABCD and BACD. Thus, we need to cancel out the duplicates by dividing 4! (there are 4! ways to shuffle the committee)

Re: combinatorics - siblings [#permalink]
16 Feb 2010, 09:11

x2suresh wrote:

bmwhype2 wrote:

There are four distinct pairs of brothers and sisters. In how many ways can a committee of 4 be formed and NOT have siblings in it?

\(= (8C1*6C1*4C1*2C1)/ 4!\) \(= 16\)

I could understand.. 8C1*6C1*4C1*2C1.. But I am not able to place.. why we divide by 4!??? Can anyone help? _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Re: There are four distinct pairs of brothers and sisters. In [#permalink]
08 Jul 2013, 16:42

What would be the ansmer if instead of a committee of 4 we would need a committee of 3? 64 possible committees?

\(2^4 * C^4_3\) = 16*4 = 64

And a committee of 2? 96? \(2^4 * C^4_2\) = 16*6 = 96 _________________

Encourage cooperation! If this post was very useful, kudos are welcome "It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James

i'm not even sure of the answer. maybe walker can help out.

4 pairs = 4*2 = 8 people total

8C4 = 8!/4!4! = 70 total outcomes

Total – unfavorable = favorable

Unfavorable outcomes Assuming one pair of twins in the committee, we have two spaces left. Since we plugged a pair of twins in the committee, we have 8-2= 6 people to fill 2 spaces.

6C2 = 6!/2!4! = 15 ways to fill the two remaining slots

We only filled the slots with one pair, and we have to account for arrangements of the pairs. Now, we have 4 pairs. 4*15= 60 total arrangements

When we place members into the remaining slots, there may be an additional set of twins. There are 2 remaining slots to which we can fit a pair of twins. If it were one remaining slot, we cannot fit a pair of twins, so we wouldn’t have to account for duplicates.

Now we account for the number of duplicates. # of duplicates = Total arrangements - # of unique combinations

To find the # of duplicates of twins, we need treat a pair of twins as one unit. This means the 4 slots are really 2 slots.

Total ways of arranging four pairs of twins in two slots 4P2 = 4*3 = 12 total ways

Total # of unique combinations Choosing two pairs out of 4 pairs = 4C2 = 6 Therefore, # of duplicates = 12 - 6 = 6 duplicates

70 – 60 + 6 = 16

Very interesting explanation. But I could not understand why we have to subtract the duplicates and how are you calculating them. _________________

Encourage cooperation! If this post was very useful, kudos are welcome "It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James

The commitee of 4 , NOT having siblings can be formed in following ways:

0 Sisters and 4 brothers = 4C4 =1 + 1 sister and 3 brothers = 4C1*3C3 = 4 [ 3C3 because selected sister's brother can not be among 3 bros] + 2 sisters and 2 brothers = 4C2* 2C2 = 6 [ again 2C2 because brothers of 2 selected sisters can not be on commitee] + 3 sisters and 1 brother = 4C3* 1C1 = 4 [ only one brother whose sister is not on commitee can be selected] + 4 sisters and 0 brothers = 4C4 = 1

Re: There are four distinct pairs of brothers and sisters. In [#permalink]
29 Apr 2015, 22:47

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Hello everyone! Researching, networking, and understanding the “feel” for a school are all part of the essential journey to a top MBA. Wouldn’t it be great... ...

Booth allows you flexibility to communicate in whatever way you see fit. That means you can write yet another boring admissions essay or get creative and submit a poem...