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

 It is currently 08 Feb 2016, 18:52

### 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
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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If there are four distinct pairs of brothers and sisters

 Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
SVP
Joined: 17 Feb 2010
Posts: 1559
Followers: 14

Kudos [?]: 391 [1] , given: 6

If there are four distinct pairs of brothers and sisters [#permalink]  28 Aug 2010, 16:14
1
This post received
KUDOS
8
This post was
BOOKMARKED
00:00

Difficulty:

15% (low)

Question Stats:

74% (01:35) correct 26% (01:13) wrong based on 421 sessions
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
Math Expert
Joined: 02 Sep 2009
Posts: 31286
Followers: 5348

Kudos [?]: 62196 [6] , given: 9444

Re: If there are four distinct pairs of brothers and sisters [#permalink]  28 Aug 2010, 16:32
6
This post received
KUDOS
Expert's post
6
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 $$C^3_4*2^3=32$$.

Answer: C.

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

Hope it helps.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 31286
Followers: 5348

Kudos [?]: 62196 [5] , given: 9444

Re: If there are four distinct pairs of brothers and sisters [#permalink]  06 Sep 2010, 12:10
5
This post received
KUDOS
Expert's post
3
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.
_________________
Manager
Joined: 30 Aug 2010
Posts: 91
Location: Bangalore, India
Followers: 4

Kudos [?]: 131 [3] , given: 27

Re: If there are four distinct pairs of brothers and sisters [#permalink]  31 Aug 2010, 23:28
3
This post received
KUDOS
5
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
Intern
Joined: 09 Oct 2009
Posts: 48
Followers: 1

Kudos [?]: 10 [2] , given: 11

Re: If there are four distinct pairs of brothers and sisters [#permalink]  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?
Intern
Joined: 10 Aug 2009
Posts: 2
Followers: 0

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

Re: If there are four distinct pairs of brothers and sisters [#permalink]  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!
Manager
Joined: 16 Mar 2010
Posts: 187
Followers: 2

Kudos [?]: 101 [0], given: 9

Re: If there are four distinct pairs of brothers and sisters [#permalink]  06 Sep 2010, 23:04
Good explaination
Math Expert
Joined: 02 Sep 2009
Posts: 31286
Followers: 5348

Kudos [?]: 62196 [0], given: 9444

Re: If there are four distinct pairs of brothers and sisters [#permalink]  07 Sep 2010, 03:11
Expert's post
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.
_________________
Intern
Joined: 09 Oct 2009
Posts: 48
Followers: 1

Kudos [?]: 10 [0], given: 11

Re: If there are four distinct pairs of brothers and sisters [#permalink]  07 Sep 2010, 04:46
Bunnuel - Thanks so much! The way you explained it was crystal clear +1!
Senior Manager
Joined: 28 Aug 2010
Posts: 254
Followers: 5

Kudos [?]: 319 [0], given: 11

Re: If there are four distinct pairs of brothers and sisters [#permalink]  23 Jan 2011, 13:04
Bunuel ....thanks a tonne.
_________________

Verbal:new-to-the-verbal-forum-please-read-this-first-77546.html
Math: new-to-the-math-forum-please-read-this-first-77764.html
Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html
-------------------------------------------------------------------------------------------------
Ajit

Senior Manager
Joined: 13 Aug 2012
Posts: 464
Concentration: Marketing, Finance
GMAT 1: Q V0
GPA: 3.23
Followers: 19

Kudos [?]: 320 [0], given: 11

Re: If there are four distinct pairs of brothers and sisters [#permalink]  27 Dec 2012, 18:38
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
$$4*8 = 32$$

Answer: C
_________________

Impossible is nothing to God.

Manager
Joined: 14 Feb 2012
Posts: 223
Followers: 2

Kudos [?]: 131 [0], given: 7

There are four distinct pairs of brothers and sisters. [#permalink]  25 Jan 2013, 00:05
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 ??/
Someone please explain ...
_________________

The Best Way to Keep me ON is to give Me KUDOS !!!
If you Like My posts please Consider giving Kudos

Shikhar

Current Student
Joined: 27 Jun 2012
Posts: 418
Concentration: Strategy, Finance
Followers: 66

Kudos [?]: 587 [0], given: 183

Re: There are four distinct pairs of brothers and sisters. [#permalink]  25 Jan 2013, 00:26
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 ??/
Someone please explain ...

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 = $$\frac{C^8_1*C^6_1*C^4_1}{3!} = \frac{8*6*4}{6}=32$$

Hence choice(C) is the answer.
_________________

Thanks,
Prashant Ponde

Tough 700+ Level RCs: Passage1 | Passage2 | Passage3 | Passage4 | Passage5 | Passage6 | Passage7
Reading Comprehension notes: Click here
VOTE: vote-best-gmat-practice-tests-excluding-gmatprep-144859.html
PowerScore CR Bible - Official Guide 13 Questions Set Mapped: Click here

Math Expert
Joined: 02 Sep 2009
Posts: 31286
Followers: 5348

Kudos [?]: 62196 [0], given: 9444

Re: There are four distinct pairs of brothers and sisters. [#permalink]  25 Jan 2013, 04:01
Expert's post
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 ??/
Someone please explain ...

Merging similar topics. Please refer to the solutions above.

This questions is also discussed here: ps-combinatorics-m02q05-55472-20.html

P.S. PLEASE SEARCH THE FORUM BEFORE POSTING.
_________________
Manager
Joined: 18 Oct 2011
Posts: 90
Location: United States
Concentration: Entrepreneurship, Marketing
GMAT Date: 01-30-2013
GPA: 3.3
Followers: 2

Kudos [?]: 48 [0], given: 0

Re: If there are four distinct pairs of brothers and sisters [#permalink]  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
Answer: C
Manager
Joined: 20 Jan 2014
Posts: 180
Location: India
Concentration: Strategy, Marketing
Followers: 0

Kudos [?]: 24 [0], given: 120

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

Consider +1 Kudos Please

Senior Manager
Joined: 15 Aug 2013
Posts: 328
Followers: 0

Kudos [?]: 34 [0], given: 23

Re: If there are four distinct pairs of brothers and sisters [#permalink]  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?
Manager
Joined: 25 Mar 2013
Posts: 95
Location: India
Concentration: Entrepreneurship, Marketing
GPA: 3.5
Followers: 1

Kudos [?]: 16 [0], given: 51

Re: If there are four distinct pairs of brothers and sisters [#permalink]  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..
VP
Joined: 08 Jul 2010
Posts: 1132
Location: India
GMAT: INSIGHT
WE: Education (Education)
Followers: 36

Kudos [?]: 814 [0], given: 40

Re: If there are four distinct pairs of brothers and sisters [#permalink]  09 Aug 2015, 05:15
Expert's post
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

Answer: option C
_________________

Prosper!!!

GMATinsight

Bhoopendra Singh and Dr.Sushma Jha

e-mail: info@GMATinsight.com
Call us : +91-9999687183 / 9891333772

http://www.GMATinsight.com/testimonials.html

Contact for One-on-One LIVE ONLINE (SKYPE Based) or CLASSROOM Quant/Verbal FREE Demo Class

READ: http://gmatclub.com/forum/620-to-760-getting-reborn-161230.html

Classroom Centre Address:
GMATinsight
107, 1st Floor, Krishna Mall,
Sector-12 (Main market),
Dwarka, New Delhi-110075

______________________________________________________
Please press the if you appreciate this post !!

Re: If there are four distinct pairs of brothers and sisters   [#permalink] 09 Aug 2015, 05:15
Similar topics Replies Last post
Similar
Topics:
8 For all positive integers f, f◎ equals the distinct pairs of 8 15 Nov 2010, 12:34
1 A code consists of either a single letter or a pair distinct 9 08 May 2010, 08:53
30 Four brothers Adam, Bill, Charles and David together 11 01 Apr 2010, 03:37
7 If there are four distinct pairs of brothers and sisters, th 8 31 Jul 2009, 22:47
19 There are four distinct pairs of brothers and sisters. In 19 02 Dec 2007, 13:57
Display posts from previous: Sort by

# If there are four distinct pairs of brothers and sisters

 Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.