It is currently 22 Nov 2017, 17:04

### 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

# Anthony and Michael sit on the six-member board of directors

Author Message
VP
Joined: 30 Jun 2008
Posts: 1031

Kudos [?]: 729 [0], given: 1

Anthony and Michael sit on the six-member board of directors [#permalink]

### Show Tags

04 Oct 2008, 06:11
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Anthony and Michael sit on the six-member board of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?

* 20%
* 30%
* 40%
* 50%
* 60%
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Kudos [?]: 729 [0], given: 1

Director
Joined: 12 Jul 2008
Posts: 514

Kudos [?]: 165 [0], given: 0

Schools: Wharton

### Show Tags

04 Oct 2008, 07:25
amitdgr wrote:
Anthony and Michael sit on the six-member board of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?

* 20%
* 30%
* 40%
* 50%
* 60%

Total # of ways to choose = 6C3 = 20
Total number of committees with both Michael and Anthony = 4*1*1 = 4

20%

Kudos [?]: 165 [0], given: 0

Intern
Joined: 04 Oct 2008
Posts: 8

Kudos [?]: 10 [1], given: 0

### Show Tags

04 Oct 2008, 09:13
1
KUDOS
The question is not "what percent of all the possible subcommittees include both Michael and Anthony?"

Then the answer would be 20%

The question is "what percent of all the possible subcommittees that include Michael also include Anthony?"

So you divide the number of possible combinations that include Michael and Anthony by the number of possible combinations that include Michael.

So, as zoinnk said there are 4 possible committees that include both Michael and Anthony.

I think it's easier to find the number of committees on which Michael does not appear at all, which would mean 5C3=10.
20-10 = 10 committees on which Michael appears.

4/10 = 40%

Kudos [?]: 10 [1], given: 0

Director
Joined: 12 Jul 2008
Posts: 514

Kudos [?]: 165 [0], given: 0

Schools: Wharton

### Show Tags

04 Oct 2008, 09:55
csvobo wrote:
The question is not "what percent of all the possible subcommittees include both Michael and Anthony?"

Then the answer would be 20%

The question is "what percent of all the possible subcommittees that include Michael also include Anthony?"

So you divide the number of possible combinations that include Michael and Anthony by the number of possible combinations that include Michael.

So, as zoinnk said there are 4 possible committees that include both Michael and Anthony.

I think it's easier to find the number of committees on which Michael does not appear at all, which would mean 5C3=10.
20-10 = 10 committees on which Michael appears.

4/10 = 40%

Sorry. Misread the question. You're right! +1

Kudos [?]: 165 [0], given: 0

VP
Joined: 30 Jun 2008
Posts: 1031

Kudos [?]: 729 [0], given: 1

### Show Tags

04 Oct 2008, 21:16
csvobo wrote:
The question is not "what percent of all the possible subcommittees include both Michael and Anthony?"

Then the answer would be 20%

The question is "what percent of all the possible subcommittees that include Michael also include Anthony?"

So you divide the number of possible combinations that include Michael and Anthony by the number of possible combinations that include Michael.

So, as zoinnk said there are 4 possible committees that include both Michael and Anthony.

I think it's easier to find the number of committees on which Michael does not appear at all, which would mean 5C3=10.
20-10 = 10 committees on which Michael appears.

4/10 = 40%

Even I did the same mistake as zoinnk ... Tricky words I guess ... zoinnk doesn't miss too many questions....

csvobo ... can you explain "to find the number of committees on which Michael does not appear at all, which would mean 5C3"

Thanks
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Kudos [?]: 729 [0], given: 1

Manager
Joined: 30 Sep 2008
Posts: 111

Kudos [?]: 20 [0], given: 0

### Show Tags

04 Oct 2008, 21:57

If Michael was sit, there is 5 positions left, in which only 2 for Michael's side. So 2/5 = 40%

Kudos [?]: 20 [0], given: 0

VP
Joined: 30 Jun 2008
Posts: 1031

Kudos [?]: 729 [0], given: 1

### Show Tags

04 Oct 2008, 22:17
lylya4 wrote:

If Michael was sit, there is 5 positions left, in which only 2 for Michael's side. So 2/5 = 40%

wow!! your method seems so much easier !!
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Kudos [?]: 729 [0], given: 1

VP
Joined: 17 Jun 2008
Posts: 1374

Kudos [?]: 426 [0], given: 0

### Show Tags

05 Oct 2008, 00:06
amitdgr wrote:
Anthony and Michael sit on the six-member board of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?

* 20%
* 30%
* 40%
* 50%
* 60%

Say there 6 members we need to form 3 member committee out of 3 members we know one is michael hence we need to select 2 out of 5
5C2 there 10 combinations

Now consider anthony and michael !!! hence select 1 out of 4 ,4C1
4 combitions

4/10 =2/5 0.4 =40%
_________________

cheers
Its Now Or Never

Kudos [?]: 426 [0], given: 0

VP
Joined: 30 Jun 2008
Posts: 1031

Kudos [?]: 729 [0], given: 1

### Show Tags

05 Oct 2008, 00:10
spriya wrote:
amitdgr wrote:
Anthony and Michael sit on the six-member board of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?

* 20%
* 30%
* 40%
* 50%
* 60%

Say there 6 members we need to form 3 member committee out of 3 members we know one is michael hence we need to select 2 out of 5
5C2 there 10 combinations

Now consider anthony and michael !!! hence select 1 out of 4 ,4C1
4 combitions

4/10 =2/5 0.4 =40%

_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Kudos [?]: 729 [0], given: 1

Re: MGMAT - Combinatorics   [#permalink] 05 Oct 2008, 00:10
Display posts from previous: Sort by

# Anthony and Michael sit on the six-member board of directors

Moderator: chetan2u

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne 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®.