Four women and three men

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

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

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Source: Beat The GMAT — Problem Solving | Sun Nov 26, 2017 5:51 am

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; 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

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8139; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

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

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