GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 14 Oct 2019, 08:57

### 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

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

# 3 men and 3 women need to be seated in 2 rows with 3 chairs

Author Message
TAGS:

### Hide Tags

Director
Joined: 07 Jun 2004
Posts: 557
Location: PA
3 men and 3 women need to be seated in 2 rows with 3 chairs  [#permalink]

### Show Tags

28 Sep 2010, 06:47
1
3
00:00

Difficulty:

25% (medium)

Question Stats:

68% (00:57) correct 32% (00:53) wrong based on 109 sessions

### HideShow timer Statistics

3 men and 3 women need to be seated in 2 rows with 3 chairs in each row. All men have to be seated in the back row . How many ways can this be done.

A. 6
B. 12
C. 24
D. 36
E. 72
Intern
Joined: 30 Mar 2010
Posts: 4

### Show Tags

28 Sep 2010, 07:01
1
_ _ _ = 3! ways women can be seated
_ _ _ = 3! ways men can be seated
3! x 3!= 36 ways total
Does anyone else agree?
Director
Joined: 07 Jun 2004
Posts: 557
Location: PA

### Show Tags

28 Sep 2010, 07:10
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
Math Expert
Joined: 02 Sep 2009
Posts: 58316

### Show Tags

28 Sep 2010, 07:22
2
1
rxs0005 wrote:
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.
_________________
Intern
Joined: 23 Apr 2009
Posts: 47
Location: Texas

### Show Tags

28 Sep 2010, 08:09
Front: 3! = 6

Back: 3! = 6

6*6=36
Non-Human User
Joined: 09 Sep 2013
Posts: 13140
Re: 3 men and 3 women need to be seated in 2 rows with 3 chairs  [#permalink]

### Show Tags

16 Jan 2019, 21:40
Hello from the GMAT Club BumpBot!

Thanks to another GMAT 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.
_________________
Re: 3 men and 3 women need to be seated in 2 rows with 3 chairs   [#permalink] 16 Jan 2019, 21:40
Display posts from previous: Sort by