combination

Expert replies

blaster: Master | Next Rank: 500 Posts; Posts: 233; Joined: Mon Jun 09, 2008 1:30 am; Thanked: 5 times

by blaster » Mon Dec 13, 2010 10:04 am

A committee of three students has to be formed. There 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 committees are possible?

Like this post

Source: Beat The GMAT — Problem Solving | Mon Dec 13, 2010 10:04 am

towerSpider: Master | Next Rank: 500 Posts; Posts: 155; Joined: Mon Dec 13, 2010 11:02 am; Thanked: 3 times

by towerSpider » Mon Dec 13, 2010 12:19 pm

blaster wrote:A committee of three students has to be formed. There 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 committees are possible?

with paul:
4
w/o paul:
4

8.

Like this post

Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Mon Dec 13, 2010 12:42 pm

blaster wrote:A committee of three students has to be formed. There 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 committees are possible?

general possibilities =5C3=10
(I) W/O Stuart => Paul-Jane-Joan; Paul-Jane-Jessica; Paul-Joan-Jessica => 3 combination sets
(II) W/O Paul => Stuart-Joan-Jessica => 1 combination set
(III) W/O Paul and Stuart => Joan-Jessica-none => 0 combination set

Combining (I, II and III) => 1+3+0=4

Last edited by Night reader on Mon Dec 13, 2010 6:26 pm, edited 1 time in total.

Like this post

diebeatsthegmat: Legendary Member; Posts: 1119; Joined: Fri May 07, 2010 8:50 amFollowed by: 3 members; Thanked: 29 times

by diebeatsthegmat » Mon Dec 13, 2010 2:23 pm

blaster wrote:A committee of three students has to be formed. There 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 committees are possible?

what the answer?
night, 40 is impossible because there is only 5!/2!/3!=10 combinations to arrange 3 people in a group include P,jane and stuart

i consider paul, stuart, jane, joan, jess = 1,2,3,4,5
1 will not be with 2 and 3 so 1 just can be with 4 and 5
1 will not be with 2 so 2 will be with 3,4,5 .consider it 4 people in a group without 1 ( paul) the combination is 4!/3!=4 ( the combination can be 2,3,4 or 235 or 245 or 345)
my answer is 5
whats the OA?

Like this post

goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

by goyalsau » Mon Dec 13, 2010 3:52 pm

diebeatsthegmat wrote:
blaster wrote:A committee of three students has to be formed. There 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 committees are possible?
what the answer?
night, 40 is impossible because there is only 5!/2!/3!=10 combinations to arrange 3 people in a group include P,jane and stuart

i consider paul, stuart, jane, joan, jess = 1,2,3,4,5
1 will not be with 2 and 3 so 1 just can be with 4 and 5
1 will not be with 2 so 2 will be with 3,4,5 .consider it 4 people in a group without 1 ( paul) the combination is 4!/3!=4 ( the combination can be 2,3,4 or 235 or 245 or 345)
my answer is 5
whats the OA?

Even did the same way and got the same answer as 5 , Just a small difference in my approach is was that i used A,B,C,D & E, for paul, stuart, jane, joan & jess respectively ..

Just wondering is there any other quick approach to solve this problem.

Saurabh Goyal
[email protected]
-------------------------

EveryBody Wants to Win But Nobody wants to prepare for Win.

Like this post

Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Mon Dec 13, 2010 6:27 pm

goyalsau wrote:
diebeatsthegmat wrote:
blaster wrote:A committee of three students has to be formed. There 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 committees are possible?
what the answer?
night, 40 is impossible because there is only 5!/2!/3!=10 combinations to arrange 3 people in a group include P,jane and stuart

i consider paul, stuart, jane, joan, jess = 1,2,3,4,5
1 will not be with 2 and 3 so 1 just can be with 4 and 5
1 will not be with 2 so 2 will be with 3,4,5 .consider it 4 people in a group without 1 ( paul) the combination is 4!/3!=4 ( the combination can be 2,3,4 or 235 or 245 or 345)
my answer is 5
whats the OA?
Even did the same way and got the same answer as 5 , Just a small difference in my approach is was that i used A,B,C,D & E, for paul, stuart, jane, joan & jess respectively ..

Just wondering is there any other quick approach to solve this problem.

I corrected the solution, as the approach was mechanical; this is simple combination with elimination

I end with 4 sets though

Like this post

prachich1987: Legendary Member; Posts: 752; Joined: Sun Sep 12, 2010 2:47 amFollowed by: 10 members; Thanked: 20 times; GMAT Score:700

by prachich1987 » Mon Dec 13, 2010 11:06 pm

blaster wrote:A committee of three students has to be formed. There 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 committees are possible?

According to me the answer is 4 teams
Plz advise whether it is correct.

The approach is as below.
Lets start with Jane

1) If Jane is there in the committee, even Paul has to be there.
Jane+Paul+third member
Third member cannot be Stuart.So the third member can be either Joan or Jessica
So the two two teams can be as below
Jane+Paul+Jessica
Jane+Paul+Joan

2) If we start forming a community with Joan, then
a) if we add Jane we will have to add hence same as 1)
b) if we add Paul then we have then we have two options for the third member--Jane/Jessica.we can't take stuart
We have already considered Jane in 1) Hence the new team can be
Joan+Paul+Jessica
c) if we add stuart then we have then we have only one option for the third member i.e Jessica.we can't take stuart. He the new team can be
Joan+Staurt+Jessica

Repeating the same procedure for remaining three members we come to the conclusion that no more new teams apart from the above 4 can be formed.
Hence the answer is 4.

Like this post

Post new topic Post Reply

• Page 1 of 1

GMAT Course Reviews

559 reviews

1168 reviews

431 reviews

359 reviews

147 reviews

5/5

5 Star (560 Reviews)

"Target Test Prep is the closest to the official version of the GMAT exam, about 99% accuracy in terms of the quality and quantity of information. The course has excellently created singular sets of focused lessons and tests for every possible topic that one could come across in the official GMAT exam."

"The TTP course maximizes the efficiency of the time you spend studying. It will take time and effort but I could almost guarantee that if you complete the course exactly as it is laid out you will get an amazing score. They also have a very responsive team willing to help with any questions you might have."

"TTP has two things that I think no other test prep company offers: A teaching approach that reinforces understanding and an attitude that will give you the mental preparedness needed to succeed on the test. TTP gives you a deep understanding of the concept you need to know while teaching you how to think."

Write a Review