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

 It is currently 18 Dec 2018, 07:58

# Expecting Soon:

R1 Admission Decisions from McCombs - Join Chat Room for Latest Updates

### 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 December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• ### Happy Christmas 20% Sale! Math Revolution All-In-One Products!

December 20, 2018

December 20, 2018

10:00 PM PST

11:00 PM PST

This is the most inexpensive and attractive price in the market. Get the course now!
• ### Key Strategies to Master GMAT SC

December 22, 2018

December 22, 2018

07:00 AM PST

09:00 AM PST

Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.

# Alice and David are two of the ten people in a group. In how many diff

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 51280
Alice and David are two of the ten people in a group. In how many diff  [#permalink]

### Show Tags

14 Aug 2018, 23:51
00:00

Difficulty:

35% (medium)

Question Stats:

75% (01:59) correct 25% (02:02) wrong based on 52 sessions

### HideShow timer Statistics

Alice and David are two of the ten people in a group. In how many different ways can this group of 10 people be divided into a group of 7 people and a group of 3 people if Alice and David are to be in the same group?

A. 48

B. 56

C. 64

D. 80

E. 128

_________________
CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2711
Location: India
GMAT: INSIGHT
WE: Education (Education)
Re: Alice and David are two of the ten people in a group. In how many diff  [#permalink]

### Show Tags

15 Aug 2018, 01:04
Bunuel wrote:
Alice and David are two of the ten people in a group. In how many different ways can this group of 10 people be divided into a group of 7 people and a group of 3 people if Alice and David are to be in the same group?

A. 48

B. 56

C. 64

D. 80

E. 128

There are two scenarios

Case 1: When Alice and David are in group of 7 : Total ways to make such groups of 7 and 3 = 8C5 (To select 5 more out of 8 to make group of 7 which includes Alice and David already)

This way the other group of 3 is automatically formed, 8C5 = 56

Case 2: When Alice and David are in group of 3 : Total ways to make such groups of 3 and 7 = 8C1 (To select 1 more out of 8 to make group of 3 which includes Alice and David already)

This way the other group of 7 is automatically formed, 8C1 = 8

Total Possible ways = 56+8 = 64

_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Intern
Joined: 28 Apr 2017
Posts: 3
Re: Alice and David are two of the ten people in a group. In how many diff  [#permalink]

### Show Tags

15 Aug 2018, 01:17
64

1*1*8c5+1*1*8c1

Sent from my XT1686 using GMAT Club Forum mobile app
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2324
Re: Alice and David are two of the ten people in a group. In how many diff  [#permalink]

### Show Tags

16 Aug 2018, 01:14
1

Solution

Given:
• There are 10 people in a group
• Alice and David are two among the ten people

To find:
• In how many ways can these 10 people be divided into a group of 7 and a group of 3, such that Alice and David are in the same group

Approach and Working:
There are two possible scenarios,
• Scenario 1: Alice and David are in the group of 7 people
• Scenario 2: Alice and David are in the group of 3 people

In Scenario 1, the total number of ways will be $$^8C_5$$
• Since, Alice and David are two among the 7 people, the other 5 people can be selected from the remaining 8 people in 8C5 ways.
• Later the remaining 3 will form another group.

In Scenario 2, the total number of ways will be $$^8C_1$$
• Since, Alice and David are two among the 3 people, the other person can be selected from the remaining 8 people in 8C1 ways.
• Later the remaining 7 will form another group.

Therefore, the total number of ways = $$^8C_5 + ^8C_1 = 8 + 56 = 64$$

Hence, the correct answer is option C.

_________________

Number Properties | Algebra |Quant Workshop

Success Stories
Guillermo's Success Story | Carrie's Success Story

Ace GMAT quant
Articles and Question to reach Q51 | Question of the week

Number Properties – Even Odd | LCM GCD | Statistics-1 | Statistics-2 | Remainders-1 | Remainders-2
Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2
Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability
Geometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry
Algebra- Wavy line | Inequalities

Practice Questions
Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets

| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com

Senior Manager
Joined: 22 Feb 2018
Posts: 414
Re: Alice and David are two of the ten people in a group. In how many diff  [#permalink]

### Show Tags

16 Aug 2018, 01:38
OA:C

No of ways, when Alice and David are in different groups $$: 2 * C(8,6) * C(2,2) = 2 *\frac{8!}{6!*2!}*1=56$$

Total number to form two teams one of $$7$$ members and another $$3$$ members, out of total $$10$$ members, without any restrictions

$$= C(10,7)*C(3,3)=\frac{10!}{7!*3!}*1=120$$

Numbers of ways when Alice and David are in the same team = Total number of ways, without restrictions - No of cases when they are in different teams

$$=120-56 = 64$$
_________________

Good, good Let the kudos flow through you

Re: Alice and David are two of the ten people in a group. In how many diff &nbs [#permalink] 16 Aug 2018, 01:38
Display posts from previous: Sort by