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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

05 Feb 2006, 22:35

6

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

39% (02:35) correct
61% (01:31) wrong based on 119 sessions

HideShow timer Statistics

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?

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

Show Tags

05 Feb 2006, 22:56

Is it 40%?

Total possibilities = 6C3 = 20

A & M can be in one in committee in 4 ways. (A,M,X where X can be any of the other 4 members)

There are 2 such committees. Hence total num of committes in which A &M are together = 4*2 = 8

Ans = 8/20 * 100 = 40%
_________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

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

Show Tags

06 Feb 2006, 06:27

all possible SUB COMS THAT INCLUDE M are 5C2=10 or 10 options.
When M and A are together then we have 4C1 or 4 options .
Then 4/10 of all subcommittees that include M also include A
So i get 40%
may be i am wrong

"# of possible to form 3 person team from 6 people = 6C3 = 20. That leaves another 20 combinations for the other team. Total = 40 possibilites (20 team A combination, 20 team B combination).

If Micheal is in team A, then Anthony has to be in team A, and team A will have only 4 ways to form (due to 4 people remaining).

But Micheal and Anthony can be in team A or team B. So % = (4/20)*2*100 = 40%"

So, we must multiply with 2!

Last edited by allabout on 06 Feb 2006, 08:08, edited 1 time in total.

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

Show Tags

06 Feb 2006, 08:07

1

This post was BOOKMARKED

Total possible combination = 6C3 = 20

Now let the members be A B C D E M
total number of 3-member-committe in which A and M are present= 4 [NOT 8]
because
(AMB , AMC, AMD, AME) is same as (MAB, MAC, MAD, MAE)

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

Show Tags

06 Feb 2006, 08:55

BG wrote:

all possible SUB COMS THAT INCLUDE M are 5C2=10 or 10 options. When M and A are together then we have 4C1 or 4 options . Then 4/10 of all subcommittees that include M also include A So i get 40% may be i am wrong

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

Show Tags

08 Jun 2015, 06:48

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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

Show Tags

08 Jun 2015, 06:49

mxr55820 wrote:

Total possible combination = 6C3 = 20

Now let the members be A B C D E M total number of 3-member-committe in which A and M are present= 4 [NOT 8] because (AMB , AMC, AMD, AME) is same as (MAB, MAC, MAD, MAE)

(4/20)*100 = 20 answer

total possible combinations = 6C3 x 2 = 40 because 2 subcom have to be formed

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? A. 20% B. 30% C. 40% D. 50% E. 60%

First approach: Let's take the group with Michael: there is a place for two other members and one of them should be taken by Anthony, as there are total of 5 people left, hence there is probability of 2/5=40%.

Second approach: Again in Michael's group 2 places are left, # of selections of 2 out of 5 5C2=10 - total # of outcomes. Select Anthony - 1C1=1, select any third member out of 4 - 4C1=4, total # =1C1*4C1=4 - total # of winning outcomes. P=# of winning outcomes/# of outcomes=4/10=40%

Third approach: Michael's group: Select Anthony as a second member out of 5 - 1/5 and any other as a third one out of 4 left 4/4, total=1/5*4/4=1/5; Select any member but Anthony as second member out of 5 - 4/5 and Anthony as a third out of 4 left 1/4, total=4/5*1/4=1/5; Sum=1/5+1/5=2/5=40%

Fourth approach: Total # of splitting group of 6 into two groups of 3: 6C3*3C3/2!=10 # of groups with Michael and Anthony: 1C1*1C1*4C1=4 P=4/10=40%

There’s something in Pacific North West that you cannot find anywhere else. The atmosphere and scenic nature are next to none, with mountains on one side and ocean on...

This month I got selected by Stanford GSB to be included in “Best & Brightest, Class of 2017” by Poets & Quants. Besides feeling honored for being part of...

Joe Navarro is an ex FBI agent who was a founding member of the FBI’s Behavioural Analysis Program. He was a body language expert who he used his ability to successfully...