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 500,000 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:

If there are four distinct pairs of brothers and sisters, then in how many ways can a committee of 3 be formed and NOT have siblings in it?

8 24 32 56 192

I find it difficult to understand the difference between permutation and combination and hence find these questions very hard.

As committee shouldn't have siblings in it, then a pair can send only one "representative" to the committee. # of ways to choose which 3 pairs of brothers and sisters should send one "representative" to the committee is \(C^3_4\) (choosing 3 pairs which will be granted the right to send one "representative" to the committee);

But each of these 3 pairs can send 2 persons to the committee either a brother or a sister: \(2*2*2=2^3\);

The first person on the committee can be anyone of the 8.

The second person on the committee can be only one out of 6 (the first person with her or his sibling excluded).

The third person can be selected only out of 4 (first two members and their siblings excluded), 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, divide the number of permutations by 6: 8*6*4 / (6) = 32 combinations.

I understand uptil the last part but I don't understand why we're diving by 6? Can anyone clarify this for me?

We need to divide \(8*6*4=192\) by the factorial of the # of people - 3! to get rid of duplications \(8*6*4=192\) contains ---> \(\frac{192}{3!}=32\) - correct answer.

Consider example with smaller number: there are two couples and we want to choose 2 people not married to each other. Couples: \(A_1\), \(A_2\) and \(B_1\), \(B_2\). Committees possible:

If we do as proposed in the solution you posted: The first person on the committee can be anyone of the 4. The second person on the committee can be only one out of 2 (the first person with her or his sibling excluded).

So we'll get: 4*2=8, so more than 4, which means that 8 contains some duplications. And to get the right answer we should divide 8 by 2! --> 8/2!=4.

It's because if you pick A1 for the first pick and than pick B1 for the second you'll get the committee {A1, B1} but if you'll pick B1 for the first pick and then A1 you'll get the exact same committee {A1, B1} (dividing by the factorial of the # of people in committee you'll exclude this double countings).

Re: If there are four distinct pairs of brothers and sisters [#permalink]

Show Tags

06 Sep 2010, 11:29

2

This post received KUDOS

The first person on the committee can be anyone of the 8.

The second person on the committee can be only one out of 6 (the first person with her or his sibling excluded).

The third person can be selected only out of 4 (first two members and their siblings excluded), 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, divide the number of permutations by 6: 8*6*4 / (6) = 32 combinations.

I understand uptil the last part but I don't understand why we're diving by 6? Can anyone clarify this for me?

Re: If there are four distinct pairs of brothers and sisters [#permalink]

Show Tags

27 Dec 2012, 18:38

1

This post received KUDOS

How many ways to select 3 of the pairs with representative in the group from 4 pairs? 4!/3!1! = 4 How many ways to select a representative from each pair? 2 x 2 x 2 = 8 \(4*8 = 32\)

AABBCCDD so ABC can come or BCD or CDA so 3! * 4 = 24 if A is not equal to A then it becomes 24*2 = 48 which ones am I missing?

Another method to solve this question is

Select any three out of these 8 individuals = 8C3 ways

Subtract the unwanted cases i.e. cases in which 2 of 3 selected have one sibling pair which can be selected as 4*6 4 = number of ways to select one sibling pair i.e. two individuals 6 = No. of ways of selecting one out of 6 remaining individuals to make a group of 3 alongwith 2 selected in previous step

Re: If there are four distinct pairs of brothers and sisters [#permalink]

Show Tags

06 Sep 2010, 17:54

Bunuel, i like the way u use quick formulas for permutations and combinations..can u let me know how can i get these formulas? i want to strengthen my skills on this subject..

Bunuel, i like the way u use quick formulas for permutations and combinations..can u let me know how can i get these formulas? i want to strengthen my skills on this subject..

Thanks!

Probability and Combinatorics chapters of Math Book:

Re: If there are four distinct pairs of brothers and sisters [#permalink]

Show Tags

11 Jul 2014, 22:42

we can select 8 people for 1st place, 6 for second (only one from pair can be selected) and , 4 for 3rd So we can have total = 8*6*4 = 192 Now in above calculation, we have counted all no of ways. (ABC is different from ABE) so we have to divide the above value with no of ways we can select 3 people

Re: If there are four distinct pairs of brothers and sisters [#permalink]

Show Tags

02 Nov 2014, 15:59

Bunuel wrote:

SnehaC wrote:

The first person on the committee can be anyone of the 8.

The second person on the committee can be only one out of 6 (the first person with her or his sibling excluded).

The third person can be selected only out of 4 (first two members and their siblings excluded), 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, divide the number of permutations by 6: 8*6*4 / (6) = 32 combinations.

I understand uptil the last part but I don't understand why we're diving by 6? Can anyone clarify this for me?

We need to divide \(8*6*4=192\) by the factorial of the # of people - 3! to get rid of duplications \(8*6*4=192\) contains ---> \(\frac{192}{3!}=32\) - correct answer.

Consider example with smaller number: there are two couples and we want to choose 2 people not married to each other. Couples: \(A_1\), \(A_2\) and \(B_1\), \(B_2\). Committees possible:

If we do as proposed in the solution you posted: The first person on the committee can be anyone of the 4. The second person on the committee can be only one out of 2 (the first person with her or his sibling excluded).

So we'll get: 4*2=8, so more than 4, which means that 8 contains some duplications. And to get the right answer we should divide 8 by 2! --> 8/2!=4.

It's because if you pick A1 for the first pick and than pick B1 for the second you'll get the committee {A1, B1} but if you'll pick B1 for the first pick and then A1 you'll get the exact same committee {A1, B1} (dividing by the factorial of the # of people in committee you'll exclude this double countings).

Hope it helps.

Hi Bunuel,

I have the same question as the other poster.

If we solve it as 8*6*4*2 == how are we creating duplicates? Aren't we eliminating the sibling by dropping down to 6 from 8 and so on?

You mention that we should divide by 2! in the above A1B1 solution. Does that mean that we would divide by 4! for the actual problem because there are four male and 4 female members or would we divide by 2 because of the sibling issue?

Re: If there are four distinct pairs of brothers and sisters [#permalink]

Show Tags

02 Dec 2014, 06:46

Combinations : Choose unique 3 from 8 ( 4 pairs) rCn 8C3 = 56 ( Total number of combinations ) Condition : Non - Siblings and Siblings !!! Non = Total - Siblings But how to find the ways of siblings?? Can anyone explain it..

If there are four distinct pairs of brothers and sisters, then in how many ways can a committee of 3 be formed and NOT have siblings in it?

A. 8 B. 24 C. 32 D. 56 E. 192

LET ABCD are Boys and PQRS are their sisters respectively Case-1: All Boys – 4C3 = 4 Case-2: All Girl – 4C3 = 4 Case-3: 2 Boys and 1 girl – 4C2*2C1 = 12 Case-4: 2 Girl and 1 Boy – 4C2*2C1 = 12

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...

Marketing is one of those functions, that if done successfully, requires a little bit of everything. In other words, it is highly cross-functional and requires a lot of different...