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

 It is currently 02 Jul 2016, 03:22

### 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

# Classic Comb/Perm/Ps

Author Message
VP
Joined: 13 Jun 2004
Posts: 1118
Location: London, UK
Schools: Tuck'08
Followers: 7

Kudos [?]: 36 [0], given: 0

### Show Tags

08 Sep 2004, 18:27
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
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4302
Followers: 34

Kudos [?]: 341 [0], given: 0

### Show Tags

08 Sep 2004, 19:07
D) 700
8c2*5c4 + 8c3*5c3 = 28*5 + 56*10 = 140 + 560 = 700
_________________

Best Regards,

Paul

Director
Joined: 20 Jul 2004
Posts: 593
Followers: 2

Kudos [?]: 93 [0], given: 0

### Show Tags

08 Sep 2004, 19:23
Paul wrote:
D) 700
8c2*5c4 + 8c3*5c3 = 28*5 + 56*10 = 140 + 560 = 700

Paul, did u consider that the two guys wont work together?

I got 700 for all possibilities, by above method. Since these two guys wont work together, the number of ways should be lesser than 700 and 635 is the only choice. E.
Director
Joined: 16 Jun 2004
Posts: 893
Followers: 2

Kudos [?]: 45 [0], given: 0

### Show Tags

08 Sep 2004, 19:37
Yep 635. It would be

6C1*2C1*5C4 + 6C2*5C4 + 6C2*2C1*5C3 + 6C3*5C3 =>60+75+300+200 = 635
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5062
Location: Singapore
Followers: 29

Kudos [?]: 282 [0], given: 0

### Show Tags

08 Sep 2004, 19:38
Combinations can be 2m,4w
3m,3w
Since the least number of men is 2, and the least number of women is 3

For 2m,4w combinations..

Number of possible outcomes for having 4 women chosen from a group of 5 women = 5C4 = 5
Number of possible outcomes for having 2 men chosen from a group of 8 men = 8C2 = 28
But 1 pair refuse to serve together, we we're left with 28-1 = 27 pairs of men

So total number of possible outcomes for 2m4w combinations = 27*5 = 135

For 3m,3w combinations...
Number of possible outcomes for having 3 women chosen from a group of 5 women = 5C3 = 10
Number of possible outcomes for having 3 men chosen from a group of 8 men = 8C3 = 56
But 1 pair of men are fickled-minded, and so we have to remove 6C1 = 6 combinations, to give 56-6 = 50 pairs of men

Total number of possible outcomes for 3m3w combinations = 50*10 = 500

Total = 500+135 = 635
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5062
Location: Singapore
Followers: 29

Kudos [?]: 282 [0], given: 0

### Show Tags

08 Sep 2004, 19:40
And as a matter of interest, if you're interested in probability, then
Number of possible outcomes = 13C6 = 1716
Number of favorable outcomes = 635

So P(at least 2 men and 3 women, 2 men refusing to serve on the same team) = 635/1716
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4302
Followers: 34

Kudos [?]: 341 [0], given: 0

### Show Tags

08 Sep 2004, 19:51
hardworker_indian wrote:
Paul wrote:
D) 700
8c2*5c4 + 8c3*5c3 = 28*5 + 56*10 = 140 + 560 = 700

Paul, did u consider that the two guys wont work together?

I got 700 for all possibilities, by above method. Since these two guys wont work together, the number of ways should be lesser than 700 and 635 is the only choice. E.

This is pretty bad... I think I skipped the last sentence...
_________________

Best Regards,

Paul

Director
Joined: 20 Jul 2004
Posts: 593
Followers: 2

Kudos [?]: 93 [0], given: 0

### Show Tags

08 Sep 2004, 19:55
Paul wrote:
hardworker_indian wrote:
Paul wrote:
D) 700
8c2*5c4 + 8c3*5c3 = 28*5 + 56*10 = 140 + 560 = 700

Paul, did u consider that the two guys wont work together?

I got 700 for all possibilities, by above method. Since these two guys wont work together, the number of ways should be lesser than 700 and 635 is the only choice. E.

This is pretty bad... I think I skipped the last sentence...

Oh oh, stop staring man. You seem to do this everyday evening in the beach.
I bet someone passed by when you skipped the last senetence.
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4302
Followers: 34

Kudos [?]: 341 [0], given: 0

### Show Tags

08 Sep 2004, 20:01
I think this is going to mark me in my GMAT quest. intr3pid what have you done to me
_________________

Best Regards,

Paul

Display posts from previous: Sort by