GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 18 Jul 2018, 13:19

Close

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
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.

Close

Request Expert Reply

Confirm Cancel

Jane and Thomas are among the 8 people from which a committee of 4 peo

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 47084
Jane and Thomas are among the 8 people from which a committee of 4 peo [#permalink]

Show Tags

New post 01 Sep 2015, 22:54
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

70% (01:31) correct 30% (02:02) wrong based on 185 sessions

HideShow timer Statistics

Jane and Thomas are among the 8 people from which a committee of 4 people is to be selected. How many different possible committees of 4 people can be selected from these 8 people if at least one of either Jane or Thomas is to be selected?

A) 28
B) 46
C) 55
D) 63
E) 70


Kudos for a correct solution.

_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

3 KUDOS received
Senior Manager
Senior Manager
User avatar
S
Joined: 21 Jan 2015
Posts: 312
Location: India
Concentration: Strategy, Marketing
GMAT 1: 620 Q48 V28
WE: Sales (Consumer Products)
GMAT ToolKit User CAT Tests
Re: Jane and Thomas are among the 8 people from which a committee of 4 peo [#permalink]

Show Tags

New post 02 Sep 2015, 00:23
3
Bunuel wrote:
Jane and Thomas are among the 8 people from which a committee of 4 people is to be selected............?

A) 28
B) 46
C) 55
D) 63
E) 70


Ans:C

Solution:
J+T+6 = total 8 ,
we need to select at least one of the J and T or both= total 4 out of 8 - 4 out of 6
8C4 - 6C4
=55
_________________

--------------------------------------------------------------------
The Mind is Everything, What we Think we Become.
Kudos will encourage many others, like me.
Please Give Kudos Image !!
Thanks :-)

3 KUDOS received
Current Student
avatar
Joined: 10 Nov 2012
Posts: 6
Location: United States
GMAT 1: 630 Q45 V31
GMAT ToolKit User
Re: Jane and Thomas are among the 8 people from which a committee of 4 peo [#permalink]

Show Tags

New post 02 Sep 2015, 01:56
3
1
No of ways to form committee with only Jane (not Thomas) : J _ _ _ ---> 6C3 = 20
No of ways to form committee with only Thomas (not Jane) : T _ _ _ ---> 6C3 = 20
No of ways to form committee with both Thomas and Jane : T J _ _ ---> 6C2 = 15

Total ways = 55
1 KUDOS received
Retired Moderator
avatar
B
Status: On a mountain of skulls, in the castle of pain, I sit on a throne of blood.
Joined: 30 Jul 2013
Posts: 348
GMAT ToolKit User Reviews Badge
Re: Jane and Thomas are among the 8 people from which a committee of 4 peo [#permalink]

Show Tags

New post 02 Sep 2015, 08:56
1
2c1*6c3 + 2c2*6c2
2*20 + 1*15
40+15
55 ways

Answer: C
1 KUDOS received
Manager
Manager
avatar
Joined: 29 Jul 2015
Posts: 159
GMAT ToolKit User
Re: Jane and Thomas are among the 8 people from which a committee of 4 peo [#permalink]

Show Tags

New post 02 Sep 2015, 09:30
1
3
Bunuel wrote:
Jane and Thomas are among the 8 people from which a committee of 4 people is to be selected. How many different possible committees of 4 people can be selected from these 8 people if at least one of either Jane or Thomas is to be selected?

A) 28
B) 46
C) 55
D) 63
E) 70


Kudos for a correct solution.


Method 1:
Number of committees in which Jane is selected is 7C3 or 35
Number of committees in which Thomas is selected is 7C3 or 35
Number of committees in which both are selected is 6C2 or 15
Number of committees in which at least one of either Jane or Thomas is selected will be

(A u B) = A + B - ( A n B)
= 35 + 35 -15
= 55

Method 2:
Total number of committees that can be formed is 8C4 or 70
Total number of committees that can be formed which do not include Jane and Thomas willl be 6C4 or 15
So, the number of committees in which at least one of either Jane or Thomas is selected will be

8C4-6C4 = 70-15
=55

Answer:- C
1 KUDOS received
Intern
Intern
avatar
Joined: 21 Jul 2015
Posts: 34
Jane and Thomas are among the 8 people from which a committee of 4 peo [#permalink]

Show Tags

New post 02 Sep 2015, 18:42
1
Bunuel wrote:
Jane and Thomas are among the 8 people from which a committee of 4 people is to be selected. How many different possible committees of 4 people can be selected from these 8 people if at least one of either Jane or Thomas is to be selected?

A) 28
B) 46
C) 55
D) 63
E) 70


Kudos for a correct solution.


Number of ways of selecting 'r' from total 'n' is given by formula ncr = n!/(r!(n-r)!)
Oder is not important in forming a committee.
Number of ways if at least one of Jane or Thomas is selected = Number of ways (Jane only is selected + Thomas only is selected + both Jane and Thomas are selected)

So, if Jane is selected in the committee, number of ways other 3 members are selected from group of 6 (8 - Jane and Thomas)
(J _ _ _)= 6C3 = 6!/(3!*(6-3)!) = 20 ways
if Thomas is selected in the committee, number of ways other 3 members are selected from group of 6 (8 - Jane and Thomas)
(T _ _ _) = 6C3 = 6!/(3!*(6-3)!) = 20 ways
if both Jane and Thomas are selected in the committee, number of ways other 2 members are selected from group of 6 (8 - Jane and Thomas)
(T J _ _) = 6C2 = 6!/(2!*(6-2)!) = 15 ways

Adding all .. 20+20+15 = 55 ways.

Answer (C)
Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 47084
Re: Jane and Thomas are among the 8 people from which a committee of 4 peo [#permalink]

Show Tags

New post 06 Sep 2015, 06:22
Bunuel wrote:
Jane and Thomas are among the 8 people from which a committee of 4 people is to be selected. How many different possible committees of 4 people can be selected from these 8 people if at least one of either Jane or Thomas is to be selected?

A) 28
B) 46
C) 55
D) 63
E) 70


Kudos for a correct solution.


KAPLAN OFFICIAL SOLUTION:

The first step in this problem is to recognize that order does not matter. We must create a committee of four people, but we are not putting these people in any sort of order. Because order does not matter, we must use the combinations formula, which is n!/[k!(n-k)!], in order to determine the number of possible outcomes.

However, as is the case with most advanced combinations problems, the solution is not as simple as plugging a few numbers into the formula. Instead, we need to consider exactly what types of outcomes we are looking for.

In this case, we need to select Jane and 3 other people, none of whom are Thomas; Thomas and 3 other people, none of whom are Jane; or Jane, Thomas and 2 other people. Additionally, we need to note that this is an ‘or’ situation, which means that we need to add all of these possibilities together.

First, if we select Jane and three others, none of whom are Thomas, we must pick Jane plus 3 out of the remaining 6 people. Because we can pick any 3 out of the 6, we set n = 6 and k = 3. When we plug this into the formula it gives us 6!/(3!3!) = 20 (Quick review: 3! = 3 x 2 x 1, 6! = 6 x 5 x 4 x 3 x 2 x 1.)

Next we determine how many ways we can select three people, none of whom are Jane, to go along with Thomas. This is identical mathematically, thus we end up with 20 in this case too.

Lastly, we determine how many outcomes are possible in which we have Jane, Thomas and 2 others. We now select 2 out of 6, so n = 6 and k = 2. 6!/(4!2!) = 15.

All we need to do to get the answer is add up all of the possible outcomes. 20 + 20 + 15 = 55, which is the correct answer to this question (C).
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Expert Post
Target Test Prep Representative
User avatar
G
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 2928
Location: United States (CA)
Re: Jane and Thomas are among the 8 people from which a committee of 4 peo [#permalink]

Show Tags

New post 31 Mar 2017, 12:36
Bunuel wrote:
Jane and Thomas are among the 8 people from which a committee of 4 people is to be selected. How many different possible committees of 4 people can be selected from these 8 people if at least one of either Jane or Thomas is to be selected?

A) 28
B) 46
C) 55
D) 63
E) 70


If at least one of either Jane or Thomas is selected for the committee, that means that either Jane is selected, Thomas is selected, or both are selected.

If Jane but not Thomas is selected, there are 6 people remaining for 3 spots. Thus, when Jane is selected, the committee can be chosen in 6C3 = (6 x 5 x 4)/3! = 20 ways. Similarly, if Thomas but not Jane is selected, the committee can be chosen in 20 ways.

Finally, if they are both selected, then there are 6 people remaining for 2 spots. Thus, the committee can be chosen in 6C2 = (6 x 5)/2! = 15 ways.

Therefore, the number of ways the committee can be chosen with at least one of either Jane or Thomas is 20 + 20 + 15 = 55 ways.

Answer: C
_________________

Scott Woodbury-Stewart
Founder and CEO

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Intern
Intern
avatar
B
Joined: 15 Aug 2017
Posts: 12
Re: Jane and Thomas are among the 8 people from which a committee of 4 peo [#permalink]

Show Tags

New post 01 Jan 2018, 05:35
can anyone help to correct me if i am wrong?

8c4 = 70 (choosing 4 out of 8)
6c4 = 15 (choosing 4 out of 6 i.e without jane or thomas)

70-15 = 55 (since either or will include jane only, thomas only and jane + thomas)...
Re: Jane and Thomas are among the 8 people from which a committee of 4 peo   [#permalink] 01 Jan 2018, 05:35
Display posts from previous: Sort by

Jane and Thomas are among the 8 people from which a committee of 4 peo

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Events & Promotions

PREV
NEXT


GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

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®.