It is currently 13 Dec 2017, 11:03

Decision(s) Day!:

CHAT Rooms | Ross R1 | Kellogg R1 | Darden R1 | Tepper R1

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

If there are four distinct pairs of brothers and sisters

Author Message
TAGS:

Hide Tags

VP
Joined: 17 Feb 2010
Posts: 1469

Kudos [?]: 805 [3], given: 6

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

Show Tags

28 Aug 2010, 16:14
3
KUDOS
21
This post was
BOOKMARKED
00:00

Difficulty:

25% (medium)

Question Stats:

70% (00:37) correct 30% (00:57) wrong based on 761 sessions

HideShow timer Statistics

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
[Reveal] Spoiler: OA

Kudos [?]: 805 [3], given: 6

Math Expert
Joined: 02 Sep 2009
Posts: 42583

Kudos [?]: 135519 [9], given: 12697

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

Show Tags

28 Aug 2010, 16:32
9
KUDOS
Expert's post
10
This post was
BOOKMARKED
seekmba wrote:
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: $$2*2*2=2^3$$;

So total # of ways is $$C^3_4*2^3=32$$.

Similar problems:
confuseddd-99055.html?hilit=married
ps-combinations-94068.html?hilit=married
combination-permutation-problem-couples-98533.html?hilit=married

Hope it helps.
_________________

Kudos [?]: 135519 [9], given: 12697

Manager
Joined: 30 Aug 2010
Posts: 91

Kudos [?]: 198 [6], given: 27

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

Show Tags

31 Aug 2010, 23:28
6
KUDOS
8
This post was
BOOKMARKED
let AABBCCDD is the group with same letter representing a sibling pair.

total # ways to select 3 from 8 is 8C3=56

Qtn: committee of 3 NOT having siblings in it = Total (56) - committee of 3 with siblings in it

committee of 3 with siblings = select 2As and one from rem. 6. This can be done in 4 ways as 4 different letters A,B,C and D
= 4 * (2C2*6C1) = 24

Hence Answer = 56-24 = 32

Kudos [?]: 198 [6], given: 27

Intern
Joined: 09 Oct 2009
Posts: 47

Kudos [?]: 29 [2], given: 11

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

Show Tags

06 Sep 2010, 11:29
2
KUDOS
1
This post was
BOOKMARKED
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?

Kudos [?]: 29 [2], given: 11

Math Expert
Joined: 02 Sep 2009
Posts: 42583

Kudos [?]: 135519 [8], given: 12697

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

Show Tags

06 Sep 2010, 12:10
8
KUDOS
Expert's post
11
This post was
BOOKMARKED
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:

$$A_1,B_1$$;
$$A_1,B_2$$;
$$A_2,B_1$$;
$$A_2,B_2$$.

Only 4 such committees are 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.
_________________

Kudos [?]: 135519 [8], given: 12697

Intern
Joined: 10 Aug 2009
Posts: 2

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

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..

Thanks!

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

Manager
Joined: 16 Mar 2010
Posts: 169

Kudos [?]: 211 [0], given: 9

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

Show Tags

06 Sep 2010, 23:04
Good explaination

Kudos [?]: 211 [0], given: 9

Math Expert
Joined: 02 Sep 2009
Posts: 42583

Kudos [?]: 135519 [0], given: 12697

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

Show Tags

07 Sep 2010, 03:11
vdixit wrote:
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:

math-probability-87244.html
math-combinatorics-87345.html

You can also see Probability and Combinatorics questions to practice at: viewforumtags.php

Hope it helps.
_________________

Kudos [?]: 135519 [0], given: 12697

Intern
Joined: 09 Oct 2009
Posts: 47

Kudos [?]: 29 [0], given: 11

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

Show Tags

07 Sep 2010, 04:46
Bunnuel - Thanks so much! The way you explained it was crystal clear +1!

Kudos [?]: 29 [0], given: 11

Senior Manager
Joined: 28 Aug 2010
Posts: 259

Kudos [?]: 806 [0], given: 11

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

Show Tags

23 Jan 2011, 13:04
Bunuel ....thanks a tonne.
_________________

Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html
-------------------------------------------------------------------------------------------------
Ajit

Kudos [?]: 806 [0], given: 11

Senior Manager
Joined: 13 Aug 2012
Posts: 457

Kudos [?]: 570 [1], given: 11

Concentration: Marketing, Finance
GPA: 3.23
Re: If there are four distinct pairs of brothers and sisters [#permalink]

Show Tags

27 Dec 2012, 18:38
1
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
$$4*8 = 32$$

_________________

Impossible is nothing to God.

Kudos [?]: 570 [1], given: 11

Current Student
Joined: 27 Jun 2012
Posts: 405

Kudos [?]: 959 [0], given: 184

Concentration: Strategy, Finance
Re: There are four distinct pairs of brothers and sisters. [#permalink]

Show Tags

25 Jan 2013, 00:26
3
This post was
BOOKMARKED
shikhar wrote:
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?

(A) 8
(B) 24
(C) 32
(D) 56
(E) 80

I got 32 with 4c3 * 2*2*2.
But why is 8c1 * 6c1 * 4c1 wrong ??/

You need to divide $$C^8_1*C^6_1*C^4_1$$with $$3!$$ to eliminate the duplicates (as the order in the arrangement does not matter).

Number of ways a committee of 3 be formed and NOT have siblings in it = $$\frac{C^8_1*C^6_1*C^4_1}{3!} = \frac{8*6*4}{6}=32$$

_________________

Thanks,
Prashant Ponde

Tough 700+ Level RCs: Passage1 | Passage2 | Passage3 | Passage4 | Passage5 | Passage6 | Passage7
VOTE GMAT Practice Tests: Vote Here
PowerScore CR Bible - Official Guide 13 Questions Set Mapped: Click here

Kudos [?]: 959 [0], given: 184

Manager
Joined: 18 Oct 2011
Posts: 89

Kudos [?]: 97 [0], given: 0

Location: United States
Concentration: Entrepreneurship, Marketing
GMAT Date: 01-30-2013
GPA: 3.3
Re: If there are four distinct pairs of brothers and sisters [#permalink]

Show Tags

25 Jan 2013, 06:41
Total number of possible committees = 56
Total number of possible committees with a sibling pair = 6C1 x 4 = 24

Therefore, total # of committees w/out a sibling pair = 56-24 = 32

Kudos [?]: 97 [0], given: 0

Current Student
Joined: 20 Jan 2014
Posts: 175

Kudos [?]: 73 [0], given: 120

Location: India
Concentration: Technology, Marketing
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

No. of ways to select 3 people = 3!

so 192/3! = 32
_________________

Kudos [?]: 73 [0], given: 120

Senior Manager
Joined: 15 Aug 2013
Posts: 298

Kudos [?]: 83 [0], given: 23

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:

$$A_1,B_1$$;
$$A_1,B_2$$;
$$A_2,B_1$$;
$$A_2,B_2$$.

Only 4 such committees are 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?

Kudos [?]: 83 [0], given: 23

Senior Manager
Joined: 25 Mar 2013
Posts: 274

Kudos [?]: 35 [0], given: 101

Location: United States
Concentration: Entrepreneurship, Marketing
GPA: 3.5
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..
_________________

I welcome analysis on my posts and kudo +1 if helpful. It helps me to improve my craft.Thank you

Kudos [?]: 35 [0], given: 101

SVP
Joined: 08 Jul 2010
Posts: 1857

Kudos [?]: 2400 [0], given: 51

Location: India
GMAT: INSIGHT
WE: Education (Education)
Re: If there are four distinct pairs of brothers and sisters [#permalink]

Show Tags

09 Aug 2015, 05:15
seekmba wrote:
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

Total Cases = 4+4+12+12 = 32

_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Kudos [?]: 2400 [0], given: 51

Senior Manager
Joined: 11 Nov 2014
Posts: 361

Kudos [?]: 55 [0], given: 17

Location: India
WE: Project Management (Telecommunications)
Re: If there are four distinct pairs of brothers and sisters [#permalink]

Show Tags

26 Sep 2016, 22:06
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?

Kudos [?]: 55 [0], given: 17

SVP
Joined: 08 Jul 2010
Posts: 1857

Kudos [?]: 2400 [0], given: 51

Location: India
GMAT: INSIGHT
WE: Education (Education)
Re: If there are four distinct pairs of brothers and sisters [#permalink]

Show Tags

26 Sep 2016, 23:40
paidlukkha wrote:
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?

Point 1: You can't take them as AABBCCDD because in sibling couple also first individual and second individuals are treated differently
Rather you should take them as A1A2 B1B2 C1C2 D1D2

In case you want to make 4 cases then
All Ones i.e. three of A1, B1, C1, D1 which can happen in 4C3 ways
All Twos i.e. three of A2, B2, C2, D2 which can happen in 4C3 ways
Two ones and one Two i.e. 4C2*2C1 = 12 (2C1 is used to select one Two out of remaining twos who are not siblings of Ones selected
Two Twos and one One i.e. 4C2*2C1 = 12 (2C1 is used to select one Two out of remaining twos who are not siblings of Ones selected

Total ways = 4+4+12+12 = 32

Point 2: you are using 3! in your solution which is completely redundant because there is no arrangement here. You only have to select 3 individuals out of 8 so the arrangement doesn't come in picture so use of 3! is completely incorrect on concept part

Point 3: I didn't understand why you used 4 in your solution.

I hope this helps!!!
_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Kudos [?]: 2400 [0], given: 51

SVP
Joined: 08 Jul 2010
Posts: 1857

Kudos [?]: 2400 [1], given: 51

Location: India
GMAT: INSIGHT
WE: Education (Education)
Re: If there are four distinct pairs of brothers and sisters [#permalink]

Show Tags

26 Sep 2016, 23:44
1
KUDOS
Expert's post
paidlukkha wrote:
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

Total favourable cases = 8C3 - (4*6) = 56 - 24 = 32
_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Kudos [?]: 2400 [1], given: 51

Re: If there are four distinct pairs of brothers and sisters   [#permalink] 26 Sep 2016, 23:44

Go to page    1   2    Next  [ 22 posts ]

Display posts from previous: Sort by