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2 men and 3 women are lined up in a row. What is the number

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by Max@Math Revolution » Thu May 26, 2016 4:24 pm
2 men and 3 women are lined up in a row. What is the number of cases where they stand with each other in turn? (The number of cases in which men (or women) do not stand next to each other)

A. 12
B. 15
C. 18
D. 21
E. 24

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by Max@Math Revolution » Sun May 29, 2016 7:50 pm
The list should be WMWMW. Hence, from women 3! And men 2!, we get (3!)(2!)=12. Therefore, the correct answer is A.
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by danielle07 » Sat Sep 02, 2017 2:36 am
This is a good one, so simple to look at. But can you give the expanded solution so i can learn how this is solved? Thanks
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by Admin1 » Sat Sep 02, 2017 1:35 pm
We have to aligned 5 people (2 men and 3 women) in a row.
The men can not stand in the first place neither the fifth place. Because if a man stands in the first place, in the second place must stand a woman, then in the third place must stand another man, in the fourth place must stand another woman and for the fifth place there is no man to pick, so we should stand a woman. Therefore, there are two women consecutive, and this is what we want to avoid.

So, the men must stand in second place and fourth place. Therefore, the women must be in the first, third and fifth place.

So we have this: _ _ _ _ _.
_
For the first place, we have to pick a woman from 3. So, we have 3 different options.

W_
For the second place, we have to pick a man from 2. So, we have 2 different options.

WM_
For the third place, we have to pick another woman from 2. So, we have 2 different options.

WMW_
For the fourth place, we have to pick the last man. So, we have 1 single option.

WMWM_
For the fifth place, we have to pick the last woman. So, we have 1 single option.

The number of cases is the product of the number of options for each place, that is to say :
3*2*2*1*1 = 12.

So the answer is A.
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by Brent@GMATPrepNow » Sun Sep 03, 2017 6:01 am
Max@Math Revolution wrote:2 men and 3 women are lined up in a row. What is the number of cases where they stand with each other in turn? (The number of cases in which men (or women) do not stand next to each other)

A. 12
B. 15
C. 18
D. 21
E. 24
NOTE: the order of the 5 people, must be: MWMWM

Take the task of arranging the 5 people and break it into stages.

Stage 1: Select a man to occupy the 1st position
There are 3 men to choose from, so we can complete stage 1 in 3 ways

Stage 2: Select a woman to occupy the 2nd position
There are 2 women to choose from, so we can complete stage 2 in 2 ways

Stage 3: Select a man to occupy the 3rd position
There are 2 men remaining to choose from, so we can complete stage 3 in 2 ways

Stage 4: Select a woman to occupy the 4th position
There is 1 woman remaining, so we can complete stage 1 in 1 way

Stage 5: Select a man to occupy the 5th position
There is 1 man remaining, so we can complete stage 1 in 1 way

By the Fundamental Counting Principle (FCP), we can complete all 5 stages (and thus arrange all 5 people) in (3)(2)(2)(1)(1) ways ([spoiler]= 12 ways[/spoiler])

Answer: A
--------------------------

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
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