Last visit was: 21 Nov 2024, 15:07 It is currently 21 Nov 2024, 15:07

Close

GRE Prep 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 GRE 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
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 12196 [19]
Given Kudos: 136
Send PM
Most Helpful Community Reply
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 12196 [5]
Given Kudos: 136
Send PM
General Discussion
avatar
Intern
Intern
Joined: 12 Aug 2018
Posts: 9
Own Kudos [?]: 6 [0]
Given Kudos: 0
Send PM
User avatar
Retired Moderator
Joined: 07 Jun 2014
Posts: 4812
Own Kudos [?]: 11188 [1]
Given Kudos: 0
GRE 1: Q167 V156
WE:Business Development (Energy and Utilities)
Send PM
Re: Four women and three men must be seated in a row for a group [#permalink]
1
Expert Reply
Runnyboy44 wrote:
May I know why is it when we place the men, we don't follow the reasoning where since there are three men, there are three men to choose from, hence 3x, followed by 2x and 1x. Instead, we follow the reasoning of the available chairs instead i.e. 5 chairs, then 4, then 3. I agree with your way as it seems correct but I can't grasp the logic behind it. Thanks for helping me to better understand.


Say there are 9 chairs so then there is a blank space to the left of a woman and a blank space to the right. Like below

_ _ _ _ _ _ _ _ _

So we can place women on chair 2 chair 4....chair 8.

Chair 2: 4 woman available for seating

Chair 4: 3 woman available for seating

Chair 6: 2 woman available for seating

Chair 8: 1 woman available for seating

So total ways = \(4 \times 3 \times 2 \times 1= 4!=24\).

Now blanks spaces available for man_1 = 5; man_2= 4; man_3=3

Total ways is \(5 \times 4 \times 3= 60\).

In the first case we are arranging the women in the second case we are arranging the blanks. When we say no two man can sit together the positions of women are fixed: i.e. 2, 4, 6, 8. Position of men are not fixed. So a viable arrangement can be:

M-W-W-M-W-M-W vs W-W-M-W-M-W-M.

If you had only 3 places for men to sit then the only viable combination would have been \(3 \times 2 \times 1= 3!\) just like the women case but here we have 5 places and 3 men, hence \(5 \times 4 \times 3=60\).


Note this is also called arranging m things in n places and it is represented by:

\(P^{n}_{m}=\frac{n!}{(n-m)!}\)

So 5 chairs 3 men: \(P^{5}_{3}=\frac{5!}{(5-3)!}= 5 \times 4 \times 3\)
avatar
Intern
Intern
Joined: 18 Jul 2018
Posts: 6
Own Kudos [?]: 2 [0]
Given Kudos: 0
Send PM
Re: Four women and three men must be seated in a row for a group [#permalink]
Why do we have to find the ways to arrange the women first? Would it be possible to solve this problem by finding all the different ways of arranging the 7 people then subtract out the number of arrangements that violate the restriction?
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 12196 [0]
Given Kudos: 136
Send PM
Re: Four women and three men must be seated in a row for a group [#permalink]
msawicka wrote:
Why do we have to find the ways to arrange the women first? Would it be possible to solve this problem by finding all the different ways of arranging the 7 people then subtract out the number of arrangements that violate the restriction?


There's only one way to find out . . . :-D
avatar
Intern
Intern
Joined: 16 Apr 2020
Posts: 1
Own Kudos [?]: 1 [0]
Given Kudos: 0
Send PM
Re: Four women and three men must be seated in a row for a group [#permalink]
1
There is another way to consider this question:
In case one: W_W_W_W
In case two: _W_W_WW
In case three: WW_W_W_

This would ensure that no two men are sitting together; it also complicates the above solution, as the solution would be 3*4!*3!, or 432.
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 12196 [0]
Given Kudos: 136
Send PM
Re: Four women and three men must be seated in a row for a group [#permalink]
Question wrote:
There is another way to consider this question:
In case one: W_W_W_W
In case two: _W_W_WW
In case three: WW_W_W_

This would ensure that no two men are sitting together; it also complicates the above solution, as the solution would be 3*4!*3!, or 432.


There are actually 10 different cases in total.
1. W_W_W_W
2. _WW_W_W
3. WW_W_W_
4. _W_WW_W
5. W_WW_W_
6. _W_W_WW
7. W_W_WW_
8. _WWW_W_
9. _W_WWW_
10. _WW_WW_

Each of the 10 possible configurations can be achieved in (4!)(3!) ways.

Cheers,
Brent
User avatar
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 5030
Own Kudos [?]: 74 [0]
Given Kudos: 0
Send PM
Re: Four women and three men must be seated in a row for a group [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Prep Club for GRE Bot
Re: Four women and three men must be seated in a row for a group [#permalink]
Moderators:
GRE Instructor
84 posts
GRE Forum Moderator
37 posts
Moderator
1111 posts
GRE Instructor
234 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne