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:

A committee of 3 students has to be formed. The are five [#permalink]
28 Jul 2007, 11:05

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

A committee of 3 students has to be formed. The are five candidates : Jane, Joan, Paul, Stuart and Jessica. If Paul and Stuart refuse to be in the committee together and Jane refuses to be in the committee without Paul, how many combinations are possible ?

A committee of 3 students has to be formed. The are five candidates : Jane, Joan, Paul, Stuart and Jessica. If Paul and Stuart refuse to be in the committee together and Jane refuses to be in the committee without Paul, how many combinations are possible ?

(i) 3 (ii) 4 (iii) 5 (iv) 6 (v) 8

Should be (ii).

If Paul is in the committee there can be 3C2 = 3 committees. (Stuart won't be there.)
I Stuart is in the committee there can only be 1 committee. (Paul and Jane won't be there.)
There can be no committee without either of them. (Stuart, Paul and Jane won't be there, so there are only 2 left.)

A committee of 3 students has to be formed. The are five candidates : Jane, Joan, Paul, Stuart and Jessica. If Paul and Stuart refuse to be in the committee together and Jane refuses to be in the committee without Paul, how many combinations are possible ?

(i) 3 (ii) 4 (iii) 5 (iv) 6 (v) 8

there are 5C3 ways to make a comittee of 3 students out of 5 w/o restriction. = 10

there are 3 ways to have a team with Paul and Stuart
there are 3 ways to have a team with Jane and Paul
There is 1 way to have a team with Jane Paul and Stewart, but it is counted twice above.

A committee of 3 students has to be formed. The are five candidates : Jane, Joan, Paul, Stuart and Jessica. If Paul and Stuart refuse to be in the committee together and Jane refuses to be in the committee without Paul, how many combinations are possible ?

(i) 3 (ii) 4 (iii) 5 (iv) 6 (v) 8

A. i would say 3.

(jean + Paul) + (either joan or jessica) = 2
(paul) + (joan and jessica) (i.e. not stuart and jean) = 1

A committee of 3 students has to be formed. The are five candidates : Jane, Joan, Paul, Stuart and Jessica. If Paul and Stuart refuse to be in the committee together and Jane refuses to be in the committee without Paul, how many combinations are possible ?

(i) 3 (ii) 4 (iii) 5 (iv) 6 (v) 8

A. i would say 3.

(jean + Paul) + (either joan or jessica) = 2 (paul) + (joan and jessica) (i.e. not stuart and jean) = 1

so total = 3

i didn't do this the formal way, but there are 4 groups

S A B
P J A
P J B
P A B

it states jean needs to be with paul, but not the other way around

A committee of 3 students has to be formed. The are five candidates : Jane, Joan, Paul, Stuart and Jessica. If Paul and Stuart refuse to be in the committee together and Jane refuses to be in the committee without Paul, how many combinations are possible ?

(i) 3 (ii) 4 (iii) 5 (iv) 6 (v) 8

A. i would say 3.

(jean + Paul) + (either joan or jessica) = 2 (paul) + (joan and jessica) (i.e. not stuart and jean) = 1

so total = 3

i didn't do this the formal way, but there are 4 groups

S A B P J A P J B P A B

it states jean needs to be with paul, but not the other way around

you are right.

i also got 4 when i used formuala but thought should use this approach to minimize the risk. hmmmmmmm fell on the trap. toooo sad!!!!!!

A committee of 3 students has to be formed. The are five candidates : Jane, Joan, Paul, Stuart and Jessica. If Paul and Stuart refuse to be in the committee together and Jane refuses to be in the committee without Paul, how many combinations are possible ?

(i) 3 (ii) 4 (iii) 5 (iv) 6 (v) 8

A. i would say 3.

(jean + Paul) + (either joan or jessica) = 2 (paul) + (joan and jessica) (i.e. not stuart and jean) = 1