PS- Committee : PS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 17 Jan 2017, 02:22

### 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

# PS- Committee

Author Message
Senior Manager
Joined: 05 Jun 2008
Posts: 307
Followers: 2

Kudos [?]: 127 [0], given: 0

### Show Tags

07 Dec 2008, 02:04
00:00

Difficulty:

(N/A)

Question Stats:

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

### HideShow timer Statistics

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

A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?

(A) 3510
(B) 2620
(C) 1404
(D) 700
(E) 635
SVP
Joined: 29 Aug 2007
Posts: 2492
Followers: 67

Kudos [?]: 734 [0], given: 19

### Show Tags

07 Dec 2008, 14:33
vivektripathi wrote:
A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?

(A) 3510
(B) 2620
(C) 1404
(D) 700
(E) 635

D.
men = 8
women = 5
total = 13
required combinations = (2 men and 4 women) + (3 men and 3 women)
required combinations = (8c2 x 5c4) + (8c3 x 5c3) = (28x5) + (56 x 10) = 700
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

Current Student
Joined: 28 Dec 2004
Posts: 3384
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 15

Kudos [?]: 281 [0], given: 2

### Show Tags

07 Dec 2008, 16:20
i think it should be 635, since 2 men dont want to serve together..

700 without such restriction
Intern
Joined: 05 Oct 2008
Posts: 14
Followers: 0

Kudos [?]: 15 [0], given: 0

### Show Tags

07 Dec 2008, 17:04
Indeed it is 700 - (5c4+5c3) = 700 - 15 = 685...

Is there a mistype or am I missing sthg?
SVP
Joined: 29 Aug 2007
Posts: 2492
Followers: 67

Kudos [?]: 734 [0], given: 19

### Show Tags

07 Dec 2008, 18:08
agree with 635.
I missed the major part - to exclude 2 men who do not share the same comittees.

= 700 - [(1 x 5c4) + (6c1 x 5c3)]
= 700 - 65
= 635
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

Senior Manager
Joined: 30 Nov 2008
Posts: 490
Schools: Fuqua
Followers: 10

Kudos [?]: 277 [0], given: 15

### Show Tags

07 Dec 2008, 20:40
I am clear till getting 700. But why did we subtract 5C4+5C3. Can you please explain.
SVP
Joined: 17 Jun 2008
Posts: 1569
Followers: 11

Kudos [?]: 250 [0], given: 0

### Show Tags

08 Dec 2008, 02:16
Somehow, I am getting a different answer. What is wrong here?
2*(7C2*5C4) + 2*(7C3*5C3) = 210 + 700 = 910
Manager
Joined: 30 Sep 2008
Posts: 111
Followers: 1

Kudos [?]: 19 [0], given: 0

### Show Tags

08 Dec 2008, 04:06
echizen wrote:
Indeed it is 700 - (5c4+5c3) = 700 - 15 = 685...

Is there a mistype or am I missing sthg?

[(8c2 -1) x 5c4] + [(8c3 -6) x 5c3] = 635

Blue: minus 1 case

Red:
Total men: A1, A2, A3, A4, A5, A6, A7, A8
If A1 & A2 don't want to serve together, as these 2 men have to be combined with one of other 6 men, so the number of the cases to be subtracted is 6
SVP
Joined: 29 Aug 2007
Posts: 2492
Followers: 67

Kudos [?]: 734 [1] , given: 19

### Show Tags

08 Dec 2008, 10:40
1
KUDOS
scthakur wrote:
Somehow, I am getting a different answer. What is wrong here?
2*(7C2*5C4) + 2*(7C3*5C3) = 210 + 700 = 910

Thats a good question and I understand what you are saying.
what you are doing is taking out one man at a time but doing so repeats some of the team that you form.
suppose m1, m2, m3, m4, m5, m6, m7, and m8 are the members of 8 men. Of them, m1 and m2 cannot be in the team. Lets consider each condition.

Condition 1: Taking out m1

Remaining members are: m2, m3, m4, m5, m6, m7, and m8, we need to combine 2 out of these 7 members. the possibilities is 7c2. fine.

Condition 2: Taking out m2

Remaining members are: m1, m3, m4, m5, m6, m7, and m8. Again we need to combine 2 men out of 7. The possibilities are same i.e. 7c2. Here the problem arises. In condition 1 you have, for example, a team of 2 consisting m3 and m4 that is repeted here as well. This way any combination from "m3, m4, m5, m6, m7, and m8" is repeted.

Therefore we cannot do => "2*(7C2*5C4) + 2*(7C3*5C3) = 210 + 700 = 910".
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

SVP
Joined: 17 Jun 2008
Posts: 1569
Followers: 11

Kudos [?]: 250 [0], given: 0

### Show Tags

08 Dec 2008, 22:40
Thanks GMATTIGER for a detailed explanation. I guess, in the beginning, I could not comprehend the repeat part. +1
Re: PS- Committee   [#permalink] 08 Dec 2008, 22:40
Display posts from previous: Sort by

# PS- Committee

 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®.