Last visit was: 24 Apr 2026, 00:40 It is currently 24 Apr 2026, 00:40
Home
Messages
Tests
Schools
Events
Close
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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
User avatar
leeye84
User avatar
Joined: 12 Mar 2007
Last visit: 08 Sep 2011
Posts: 138
Own Kudos:
1,734
Given Kudos: 2
Posts: 138
Kudos: 1,734
There are four distinct paris of brothers and sisters. In 13 Jul 2007, 14:55
Kudos
Add Kudos
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
N
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

History

Date
Time
Result
Not Attempted Yet
There are four distinct paris 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



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
User avatar
dahcrap
User avatar
Joined: 03 Jun 2007
Last visit: 27 May 2008
Posts: 159
Own Kudos:
33
Posts: 159
Kudos: 33
There are four distinct paris of brothers and sisters. In 13 Jul 2007, 15:20
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I get 32. Today am getting many problems wrong so I hope to redeem myself with this one :-(

How 32 ? The committee can be all 3 boys or all 3 girls. For this we 2 *4C3 = 8
Next the committee can have 2 boys and 1 girl or 2 girls and 1 boys
For this
4C2 * 2 + 4C2*2 = 12 + 12 = 24.24+8 = 32
User avatar
briks123
User avatar
Joined: 08 Jul 2007
Last visit: 04 Aug 2007
Posts: 66
Own Kudos:
31
Posts: 66
Kudos: 31
There are four distinct paris of brothers and sisters. In 13 Jul 2007, 15:41
Kudos
Add Kudos
Bookmarks
Bookmark this Post
dahcrap
I get 32. Today am getting many problems wrong so I hope to redeem myself with this one :-(

How 32 ? The committee can be all 3 boys or all 3 girls. For this we 2 *4C3 = 8
Next the committee can have 2 boys and 1 girl or 2 girls and 1 boys
For this
4C2 * 2 + 4C2*2 = 12 + 12 = 24.24+8 = 32
Show more


I agree with you this time dahcrap! Here's my explanation: There are 8!/(3!5!)=56 ways to form the groups without the restriction. In these 56 ways, there are 24 times that siblings are in a group. (If A and B are siblings, then there are 6 other people who can be paired with them to form a group. There are four pairs of siblings so 4*6=24). So 56-24=32!
User avatar
boubi
User avatar
Joined: 24 Jun 2006
Last visit: 09 Jul 2016
Posts: 27
Own Kudos:
82
Products:
My Reviews
Posts: 27
Kudos: 82
There are four distinct paris of brothers and sisters. In 13 Jul 2007, 15:45
Kudos
Add Kudos
Bookmarks
Bookmark this Post
i don't know dude maybe you are right... i just rely on my intuitive skills for this...maybe can i explain my reasonning.
first 8C3 is the total way to choose a group of 3 among 8 people
then i substract the way you can choose a team of three withou siblings thus the first in the group will have 8 possibilities 8C1 then the second will have only 6 possiblities (we substract their siblings that cannot be part of the group) 6C1 and so on to 4C1.
however we hav implied an order in multipliyng the C's and we don't want one, so i divide by 3!....

hope i am right...
can you explain your reasoning?
User avatar
briks123
User avatar
Joined: 08 Jul 2007
Last visit: 04 Aug 2007
Posts: 66
Own Kudos:
31
Posts: 66
Kudos: 31
There are four distinct paris of brothers and sisters. In 13 Jul 2007, 16:08
Kudos
Add Kudos
Bookmarks
Bookmark this Post
boubi
i don't know dude maybe you are right... i just rely on my intuitive skills for this...maybe can i explain my reasonning.
first 8C3 is the total way to choose a group of 3 among 8 people
then i substract the way you can choose a team of three withou siblings thus the first in the group will have 8 possibilities 8C1 then the second will have only 6 possiblities (we substract their siblings that cannot be part of the group) 6C1 and so on to 4C1.
however we hav implied an order in multipliyng the C's and we don't want one, so i divide by 3!....

hope i am right...
can you explain your reasoning?
Show more


Not sure I understand what you mean...

But here's what I was thinking:

Let the siblings be the following:
AB
CD
EF
GH

We are obviously in agreement that there are 56 ways to make a group of 3 without the sibling restriction. I think it is easier to count how many of those 56 ways have siblings in them.

So, when A and B are in a group, C, D, E, F, G, and H can also be matched with them. So 6 of those ways have A and B in the same group. The same logic can be applied for the other three pairs. When C and D are in a group, A, B, E, F, G, and H can be matched with them to make the group of three. SO 6*4 because there are four pairs of siblings is 24 groups in which there are siblings. So there are 56-24 groups without siblings.

Does this make sense? Maybe I'm wrong...
User avatar
boubi
User avatar
Joined: 24 Jun 2006
Last visit: 09 Jul 2016
Posts: 27
Own Kudos:
82
Products:
My Reviews
Posts: 27
Kudos: 82
There are four distinct paris of brothers and sisters. In 13 Jul 2007, 16:42
Kudos
Add Kudos
Bookmarks
Bookmark this Post
this make perfect sense!! seeing all those C's make me struggle between crazymath and common sense logic.
the worst is that i ve already use your way to solve this problem...

know let's figure out what is wrong with my band of C's.??!! :)
User avatar
Himalayan
User avatar
Joined: 26 Feb 2006
Last visit: 09 Aug 2011
Posts: 383
Own Kudos:
633
Posts: 383
Kudos: 633
There are four distinct paris of brothers and sisters. In 13 Jul 2007, 17:50
Kudos
Add Kudos
Bookmarks
Bookmark this Post
leeye84
There are four distinct paris 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
Show more


= total - a combination of a sibling + a non sibling
= 8c3 - (4)(6c1)
= 54 - 24
= 32



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
Moderator:
Bunuel
Math Expert
109802 posts