GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Aug 2019, 11:36

### 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

# If there are four distinct pairs of brothers and sisters

Author Message
TAGS:

### Hide Tags

Manager
Joined: 12 Nov 2016
Posts: 135
Concentration: Entrepreneurship, Finance
GMAT 1: 620 Q36 V39
GMAT 2: 650 Q47 V33
Re: If there are four distinct pairs of brothers and sisters  [#permalink]

### Show Tags

04 Nov 2017, 11:25
Bunuel wrote:
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 $$C^3_4*2^3=32$$.

What I fail to understand in this question and a similar one with 5 married couples, is a second part - why a number of alternatives for 3 pairs to send 2 persons equal to 2*2*2= 8
Here is a small example, let pairs be AB, CD, EF. Since 2 persons can't form the original pair of siblings then the alternatives are:
1)AC 2)AD 3) AE 4) AF 5) BC 6) BD 7) BE 8) BF 9) CE 10) CF 11) DE 12) DF
So, the number of alternatives is 12, not 8! What is it that I am missing?
Director
Joined: 14 Dec 2017
Posts: 517
Location: India
Re: If there are four distinct pairs of brothers and sisters  [#permalink]

### Show Tags

11 Jun 2018, 02:57
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

# of ways to from a committee of 3 with no siblings on it = (8 * 6 * 4)/3! = 32

The First person in the committee can be chosen in 8 ways.
The Second person in the committee can be chosen in 6 ways, since we will remove the sibling of first person selected & then choose.
The Third person in the committee can be chosen in 4 ways, since we will remove the sibling of second person selected, as well & then choose.

Now we are asked for the # of ways to form a committee & not the arrangement of the persons, hence order does not matter. To account for this we divide by 3!, which is the # of possible arrangements of 3 persons.

Thanks,
GyM
_________________
Intern
Joined: 09 Feb 2019
Posts: 1
Re: If there are four distinct pairs of brothers and sisters  [#permalink]

### Show Tags

15 Jul 2019, 23:16
I know the answer is 32. But what's the difference between the 2 approaches?

Case 1: selected any 3 pairs of bro-sis and then one from each pair:
4C3 X 2C1 X 2C1 X 2C1

Case 2: Selected 1 person out of each pair and then 3 out of those people
2C1 X 2C1 X 2C1 X 2C1 X 4C3

Posted from my mobile device
Re: If there are four distinct pairs of brothers and sisters   [#permalink] 15 Jul 2019, 23:16

Go to page   Previous    1   2   [ 23 posts ]

Display posts from previous: Sort by