Last visit was: 24 Apr 2026, 14:44 It is currently 24 Apr 2026, 14:44
Close
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
Your Progress

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
rxs0005
Joined: 07 Jun 2004
Last visit: 21 Jun 2017
Posts: 436
Own Kudos:
3,310
 [11]
Given Kudos: 22
Location: PA
Posts: 436
Kudos: 3,310
 [11]
3
Kudos
Add Kudos
8
Bookmarks
Bookmark this Post
avatar
Bruno2000
Joined: 30 Mar 2010
Last visit: 10 Nov 2010
Posts: 3
Own Kudos:
1
 [1]
Given Kudos: 4
Posts: 3
Kudos: 1
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
rxs0005
Joined: 07 Jun 2004
Last visit: 21 Jun 2017
Posts: 436
Own Kudos:
Given Kudos: 22
Location: PA
Posts: 436
Kudos: 3,310
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 24 Apr 2026
Posts: 109,818
Own Kudos:
811,068
 [4]
Given Kudos: 105,873
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,818
Kudos: 811,068
 [4]
3
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
rxs0005
I agree

what i had a doubt was since men have to be in the back row thats 3! ways and because of this women will be in the front row thats 3! why cant it be 6 + 6 = 12 ways since the row restriction is already in place

Principle of Multiplication
If one event can occur in \(m\) ways and a second can occur independently of the first in \(n\) ways, then the two events can occur in \(mn\) ways.

Or consider this: for one particular arrangement of men, say {m1, m2, m3} women in the front row can be arranged in 3!=6 ways, as total # of arrangements of men is 3!=6 then total # of arrangements of men and women is 3!*3!=36.

Hope it's clear.
User avatar
Lethal
Joined: 23 Apr 2009
Last visit: 04 Mar 2011
Posts: 42
Own Kudos:
Given Kudos: 4
Location: Texas
Posts: 42
Kudos: 6
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Front: 3! = 6

Back: 3! = 6

6*6=36
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,974
Own Kudos:
Posts: 38,974
Kudos: 1,117
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Math Expert
109818 posts
Tuck School Moderator
853 posts