It is currently 26 Jun 2017, 22:42

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

GMATprep practice question: can somebody please solve it

Author Message
Manager
Joined: 05 Jan 2009
Posts: 81

Show Tags

25 Apr 2009, 15:43
00:00

Difficulty:

(N/A)

Question Stats:

50% (02:14) correct 50% (00:00) wrong based on 3 sessions

HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Could you please solve it with explanation in each step?

A committee of three people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee:

A. 16
B. 24
C. 26
D. 30
E. 32

[Reveal] Spoiler:
OA is 32
Intern
Joined: 18 Feb 2009
Posts: 5

Show Tags

25 Apr 2009, 22:33
I got E.

since no married couple is allowed, consider 4 couples as 4 element. So, we need to select the 3 couples from 4 and then who among the selected couple will be part of the group.

3 couple can be selected in 4c3 ways and a member can be selected in 2 ways from each of the selected couple.

So, total number of ways will be = 4c3*2*2*2 = 32

Thanks
Akshay
Manager
Joined: 05 Jan 2009
Posts: 81

Show Tags

26 Apr 2009, 14:53
Why are you doing 2*2*2? can you please explain?
Manager
Joined: 22 Jul 2009
Posts: 191

Show Tags

21 Sep 2009, 13:37
. Four married couples = 8 people
. 3 spots

8 * 6 * 4 = number of possible permutations

But here order does not matter => we want the number of possible combinations, not permutations.

nPk = nCk * k!

=> 8 * 6 * 4 / 3! = 32
_________________

Please kudos if my post helps.

Last edited by powerka on 22 Sep 2009, 06:16, edited 1 time in total.
Manager
Joined: 22 Jul 2009
Posts: 191

Show Tags

21 Sep 2009, 13:40
pmal04 wrote:
Why are you doing 2*2*2? can you please explain?

4C3 * 2! * 2! * 2! is an alternative approach, equally valid.

That's picking 3 people out of 4 couples, and then accounting for the possible combinations within each couple (each couple has two people, so possible combinations within each couple are 2!).
_________________

Please kudos if my post helps.

Intern
Joined: 02 Sep 2009
Posts: 8

Show Tags

21 Sep 2009, 14:28
I got E as well.

My answer may not be as efficient as some but here is what I did.

Step 1: I started with 8 Choose 3 - since we have a total of 8 and are picking a group of 3. Choose Combination instead of Permutations since order does not matter. 8 Choose 3 = 56

Step 2: I know that x<56 since some of the combinations above will yield 2 people from the same married group. So next I figured out how many combinations would yield a married couple in the group. I thought of the population as follows to do so quickly;

A1 A2 B1 B2 C1 C2 D1 D2 -

If there is a married couple in 1 group and only 1 spot left I deduced that it could only be 6 different people in with each married couple. So 4 married couples x 6 available options yields 24 groups which have a married couple (6x4=24)

Step 3: 56 - 24 = 32
Re: GMATprep practice question: can somebody please solve it   [#permalink] 21 Sep 2009, 14:28
Display posts from previous: Sort by