BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Four women and three men

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Moderator
Posts: 7187
Joined: Thu Sep 07, 2017 4:43 pm
Followed by:23 members

Four women and three men

by BTGmoderatorDC » Sun Nov 26, 2017 5:51 am
Four women and three men must be seated in a row for a group photograph. If no two men can sit next to each other, in how many different ways can the seven people be seated?

A) 240
B) 480
C) 720
D) 1440
E) 5640

What is the easiest way to solve this kind of problem?

OA D
Top
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Sun Nov 26, 2017 7:21 am
lheiannie07 wrote:Four women and three men must be seated in a row for a group photograph. If no two men can sit next to each other, in how many different ways can the seven people be seated?

A) 240
B) 480
C) 720
D) 1440
E) 5640

What is the easiest way to solve this kind of problem?

OA D
Take the task of arranging the 7 peopl and break it into stages.

Stage 1: Arrange the 4 women in a row
We can arrange n unique objects in n! ways.
So, we can arrange the 4 women in 4! ways (= 24 ways)
So, we can complete stage 1 in 24 ways

IMPORTANT: For each arrangement of 4 women, there are 5 spaces where the 3 men can be placed.
If we let W represent each woman, we can add the spaces as follows: _W_W_W_W_
So, if we place the men in 3 of the available spaces, we can ENSURE that two men are never seated together.

Let's let A, B and C represent the 3 men.

Stage 2: Place man A in an available space.
There are 5 spaces, so we can complete stage 2 in 5 ways.

Stage 3: Place man B in an available space.
There are 4 spaces remaining, so we can complete stage 3 in 4 ways.

Stage 4: Place man C in an available space.
There are 3 spaces remaining, so we can complete stage 4 in 3 ways.

By the Fundamental Counting Principle (FCP), we can complete the 4 stages (and thus seat all 7 people) in (24)(5)(4)(3) ways (= 1440 ways)

Answer: D

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat- ... /video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 8083
Joined: Sat Apr 25, 2015 10:56 am
Location: Los Angeles, CA
Thanked: 43 times
Followed by:29 members

by Scott@TargetTestPrep » Mon Oct 07, 2019 10:21 am
BTGmoderatorDC wrote:Four women and three men must be seated in a row for a group photograph. If no two men can sit next to each other, in how many different ways can the seven people be seated?

A) 240
B) 480
C) 720
D) 1440
E) 5640

What is the easiest way to solve this kind of problem?

OA D
We can see that we have to use the 4 women to separate the 3 men.

_W_W_W_W_

From the above, we see that if we place each of the 3 men in any of the 5 blanks above, we have separated the 3 men. So there are 5C3 = (5 x 4 x 3)/(3 x 2) = 10 ways to separate the 3 men. Now, once we have a 7-person lineup, there are 4! x 3! = 24 x 6 = 144 ways to arrange them (4! for the women and 3! for the men). Therefore, in total, we have 10 x 144 = 1440 seating arrangements such that no two men are sitting next to each other.

Answer: D

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage
Top
Post new topic Post Reply
• Page 1 of 1