Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 25 May 2017, 14:22

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# A group of four women and three men have tickets for seven a

Author Message
TAGS:

### Hide Tags

Intern
Joined: 23 Dec 2009
Posts: 47
Schools: HBS 2+2
WE 1: Consulting
WE 2: Investment Management
Followers: 1

Kudos [?]: 42 [3] , given: 7

A group of four women and three men have tickets for seven a [#permalink]

### Show Tags

30 Dec 2009, 21:43
3
KUDOS
7
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

62% (02:05) correct 38% (01:25) wrong based on 437 sessions

### HideShow timer Statistics

A group of four women and three men have tickets for seven adjacent seats in one row of a theatre. If the three men will not sit in three adjacent seats, how many possible different seating arrangements are there for these 7 theatre-goers?

(A) 7! – 2!3!2!
(B) 7! – 4!3!
(C) 7! – 5!3!
(D) 7 × 2!3!2!
(E) 2!3!2!
[Reveal] Spoiler: OA

_________________

My GMAT quest...

...over!

Manager
Joined: 09 May 2009
Posts: 203
Followers: 1

Kudos [?]: 236 [4] , given: 13

Re: Combinations Problem -- Arrangement of Seats [#permalink]

### Show Tags

31 Dec 2009, 00:31
4
KUDOS
IMO C

7 people can be seated in 7! ways

take 3 men as one unit ----> tot 5 people can be seated in 5 ways *(no. of ways in which 4 women can be seated amng themselves ) * ( no. of ways in which 3 men cen be seated amng themselves)=5*4!*3!=5!*3!

tot no. of ways in which 3 men are not seated in adjacent seats=tot arrangements - 5!*3!=7!-5!*3!
_________________

GMAT is not a game for losers , and the moment u decide to appear for it u are no more a loser........ITS A BRAIN GAME

Manager
Joined: 30 Jun 2004
Posts: 177
Location: Singapore
Followers: 1

Kudos [?]: 24 [0], given: 5

Re: Combinations Problem -- Arrangement of Seats [#permalink]

### Show Tags

31 Dec 2009, 01:33
1
This post was
BOOKMARKED
I agree, the answer sems to be C.

7 people can sit in 7! different ways. But because 3 men cannot sit together, we take them as a unit.

This unit of men, among themselves can sit in 3! ways.

Hence, 7! - 3!.

This unit of men along with 4 women can sit in 5! different ways which also needs to be eliminated.

Hence 7! - 5!3!
Intern
Joined: 23 Dec 2009
Posts: 47
Schools: HBS 2+2
WE 1: Consulting
WE 2: Investment Management
Followers: 1

Kudos [?]: 42 [0], given: 7

Re: Combinations Problem -- Arrangement of Seats [#permalink]

### Show Tags

31 Dec 2009, 03:35
xcusemeplz2009 wrote:
IMO C

7 people can be seated in 7! ways

take 3 men as one unit ----> tot 5 people can be seated in 5 ways *(no. of ways in which 4 women can be seated amng themselves ) * ( no. of ways in which 3 men cen be seated amng themselves)=5*4!*3!=5!*3!

tot no. of ways in which 3 men are not seated in adjacent seats=tot arrangements - 5!*3!=7!-5!*3!

I understand having 7! total arrangements and subtracting out 4!3!, but why do why multiply this term we subtract out, 4!3! by 5? Is it because there are 5 situations where 3 men are next to each other (see below)?

1: MMMWWWW
2: WMMMWWW
3: WWMMMWW
4: WWWMMMW
5: WWWWMMM
_________________

My GMAT quest...

...over!

Math Expert
Joined: 02 Sep 2009
Posts: 38880
Followers: 7733

Kudos [?]: 106123 [5] , given: 11607

Re: Combinations Problem -- Arrangement of Seats [#permalink]

### Show Tags

31 Dec 2009, 19:53
5
KUDOS
Expert's post
11
This post was
BOOKMARKED
A group of four women and three men have tickets for seven adjacent seats in one row of a theatre. If the three men will not sit in three adjacent seats, how many possible different seating arrangements are there for these 7 theatre-goers?

(A) 7! – 2!3!2!
(B) 7! – 4!3!
(C) 7! – 5!3!
(D) 7 × 2!3!2!
(E) 2!3!2!

There are 3 men and 4 women, we want to calculate the seating arrangements if three men do not sit together, like MMM.

Let's calculate the # of arrangements when they SIT together and subtract from total # of arrangements of these 7 persons without restriction. Thus we'll get the # of arrangements asked in the question.

1. Total # of arrangements of 7 is 7!.

2. # of arrangements when 3 men are seated together, like MMM;

Among themselves these 3 men can sit in 3! # of ways,
Now consider these 3 men as one unit like this {MMM}. We'll have total of 5 units: {MMM}{W}{W}{W}{W}. The # of arrangements of these 5 units is 5!.

Hence total # of arrangements when 3 men sit together is: 3!5!.

# of arrangements when 3 men do not sit together would be: 7!-3!5!.

Hope it's clear.
_________________
Senior Manager
Joined: 22 Dec 2009
Posts: 359
Followers: 11

Kudos [?]: 389 [0], given: 47

Re: Combinations Problem -- Arrangement of Seats [#permalink]

### Show Tags

01 Jan 2010, 10:07

Total arrangments posb = 7!

Treat 3 Men as a single unit. Hence Men + 4 women can be arranged in 5 ways.
3 Men within the single unit can be arranged in 3! ways
4 women can be arranged in 4! ways.

Therefore no of posb when 3 men sit adjacent to each other (as a single unit) = 5x3!x4! = 5! x 3!

Hence no of posb when 3 men dont sit together = 7! - 5! x 3!

Cheers!
JT
_________________

Cheers!
JT...........
If u like my post..... payback in Kudos!!

|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~Better Burn Out... Than Fade Away~~

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15444
Followers: 649

Kudos [?]: 209 [0], given: 0

Re: A group of four women and three men have tickets for seven a [#permalink]

### Show Tags

21 Sep 2013, 12:40
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.
_________________
Manager
Status: Time to wait :)
Joined: 18 Jun 2013
Posts: 80
Location: India
Concentration: Entrepreneurship, Strategy
GMAT 1: 710 Q48 V40
WE: Consulting (Telecommunications)
Followers: 5

Kudos [?]: 52 [1] , given: 1181

Re: Combinations Problem -- Arrangement of Seats [#permalink]

### Show Tags

21 Sep 2013, 12:46
1
KUDOS
Bunuel wrote:
A group of four women and three men have tickets for seven adjacent seats in one row of a theatre. If the three men will not sit in three adjacent seats, how many possible different seating arrangements are there for these 7 theatre-goers?

(A) 7! – 2!3!2!
(B) 7! – 4!3!
(C) 7! – 5!3!
(D) 7 × 2!3!2!
(E) 2!3!2!

There are 3 men and 4 women, we want to calculate the seating arrangements if three men do not sit together, like MMM.

Let's calculate the # of arrangements when they SIT together and subtract from total # of arrangements of these 7 persons without restriction. Thus we'll get the # of arrangements asked in the question.

1. Total # of arrangements of 7 is 7!.

2. # of arrangements when 3 men are seated together, like MMM;

Among themselves these 3 men can sit in 3! # of ways,
Now consider these 3 men as one unit like this {MMM}. We'll have total of 5 units: {MMM}{W}{W}{W}{W}. The # of arrangements of these 5 units is 5!.

Hence total # of arrangements when 3 men sit together is: 3!5!.

# of arrangements when 3 men do not sit together would be: 7!-3!5!.

Hope it's clear.

You tend to write "Hope it's clear." after every solution, but there "never is" a chance that you have explained something and it isn't clear. Unlimited Kudos to you, and RESPECT!
_________________

Dream On...

Current Student
Joined: 06 Sep 2013
Posts: 2005
Concentration: Finance
Followers: 68

Kudos [?]: 643 [0], given: 355

Re: A group of four women and three men have tickets for seven a [#permalink]

### Show Tags

29 Dec 2013, 17:18
R2I4D wrote:
A group of four women and three men have tickets for seven adjacent seats in one row of a theatre. If the three men will not sit in three adjacent seats, how many possible different seating arrangements are there for these 7 theatre-goers?

(A) 7! – 2!3!2!
(B) 7! – 4!3!
(C) 7! – 5!3!
(D) 7 × 2!3!2!
(E) 2!3!2!

7 people can be seated in 7!

Now, we need to plot the unfavorable scenario, that is 3 men sit together

Group them as per glue method as one entity. Now we have to arrange the 5!
Within the group of 3 men they can be arranged in 3!

So total number of arrangements is 5!3!

Now favorable scenario will be = Total - unfavorable

So total is 7! - 5!3!

Hope it helps
Cheers!
J
Intern
Joined: 23 Aug 2013
Posts: 46
Followers: 0

Kudos [?]: 10 [0], given: 8

Re: Combinations Problem -- Arrangement of Seats [#permalink]

### Show Tags

11 Jun 2014, 21:52
Bunuel wrote:
A group of four women and three men have tickets for seven adjacent seats in one row of a theatre. If the three men will not sit in three adjacent seats, how many possible different seating arrangements are there for these 7 theatre-goers?

(A) 7! – 2!3!2!
(B) 7! – 4!3!
(C) 7! – 5!3!
(D) 7 × 2!3!2!
(E) 2!3!2!

There are 3 men and 4 women, we want to calculate the seating arrangements if three men do not sit together, like MMM.

Let's calculate the # of arrangements when they SIT together and subtract from total # of arrangements of these 7 persons without restriction. Thus we'll get the # of arrangements asked in the question.

1. Total # of arrangements of 7 is 7!.

2. # of arrangements when 3 men are seated together, like MMM;

Among themselves these 3 men can sit in 3! # of ways,
Now consider these 3 men as one unit like this {MMM}. We'll have total of 5 units: {MMM}{W}{W}{W}{W}. The # of arrangements of these 5 units is 5!.

Hence total # of arrangements when 3 men sit together is: 3!5!.

# of arrangements when 3 men do not sit together would be: 7!-3!5!.

Hope it's clear.

A silly doubt that have cropped up all of a sudden

Bunuel, I've a doubt. Why are we not dividing 5! by 4! as there are 4 of the same type in the group. I know I'm wrong. Kindly help me where
Math Expert
Joined: 02 Sep 2009
Posts: 38880
Followers: 7733

Kudos [?]: 106123 [0], given: 11607

Re: Combinations Problem -- Arrangement of Seats [#permalink]

### Show Tags

12 Jun 2014, 04:53
sgangs wrote:
Bunuel wrote:
A group of four women and three men have tickets for seven adjacent seats in one row of a theatre. If the three men will not sit in three adjacent seats, how many possible different seating arrangements are there for these 7 theatre-goers?

(A) 7! – 2!3!2!
(B) 7! – 4!3!
(C) 7! – 5!3!
(D) 7 × 2!3!2!
(E) 2!3!2!

There are 3 men and 4 women, we want to calculate the seating arrangements if three men do not sit together, like MMM.

Let's calculate the # of arrangements when they SIT together and subtract from total # of arrangements of these 7 persons without restriction. Thus we'll get the # of arrangements asked in the question.

1. Total # of arrangements of 7 is 7!.

2. # of arrangements when 3 men are seated together, like MMM;

Among themselves these 3 men can sit in 3! # of ways,
Now consider these 3 men as one unit like this {MMM}. We'll have total of 5 units: {MMM}{W}{W}{W}{W}. The # of arrangements of these 5 units is 5!.

Hence total # of arrangements when 3 men sit together is: 3!5!.

# of arrangements when 3 men do not sit together would be: 7!-3!5!.

Hope it's clear.

A silly doubt that have cropped up all of a sudden

Bunuel, I've a doubt. Why are we not dividing 5! by 4! as there are 4 of the same type in the group. I know I'm wrong. Kindly help me where

All men and women are different, so no need for factorial correction there. For example, arrangement {Bill, Bob, Ben} {Ann}, {Beth}, {Carol}, {Diana} is different from {Bill, Bob, Ben}, {Beth}, {Carol}, {Diana}, {Ann}.

Hope it's clear.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15444
Followers: 649

Kudos [?]: 209 [0], given: 0

Re: A group of four women and three men have tickets for seven a [#permalink]

### Show Tags

28 Jul 2015, 20:16
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.
_________________
CEO
Joined: 17 Jul 2014
Posts: 2502
Location: United States (IL)
Concentration: Finance, Economics
Schools: Stanford '19 (D)
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
Followers: 26

Kudos [?]: 342 [0], given: 168

Re: A group of four women and three men have tickets for seven a [#permalink]

### Show Tags

10 Feb 2016, 19:17
R2I4D wrote:
A group of four women and three men have tickets for seven adjacent seats in one row of a theatre. If the three men will not sit in three adjacent seats, how many possible different seating arrangements are there for these 7 theatre-goers?

(A) 7! – 2!3!2!
(B) 7! – 4!3!
(C) 7! – 5!3!
(D) 7 × 2!3!2!
(E) 2!3!2!

you can get to the answer choice by applying logic.

1. we have 7 seats, to technically, without restrictions, we would have 7! combinations. From 7!, we would extract the number of combinations in which the men are together.
Right away, we can eliminate D and E.

since the order does matter, we need to use combinations:
suppose all the guys are 1 single guy.
thus, we would have 4W and 1M.
we can arrange 1 guy and 4w in 5 ways.
thus, we would have 5!
since the number of combinations would be greater..since no two guys must be alone, it must be true that the number of combinations in which at least some 2 guys are near each other should be greater than 5!X, where x is a coefficient.
we can eliminate A and B right away, since neither of them would give something at least closer to 5...
C
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15444
Followers: 649

Kudos [?]: 209 [0], given: 0

Re: A group of four women and three men have tickets for seven a [#permalink]

### Show Tags

13 Feb 2017, 11:23
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: A group of four women and three men have tickets for seven a   [#permalink] 13 Feb 2017, 11:23
Similar topics Replies Last post
Similar
Topics:
23 Four women and three men must be seated in a row for a group photo 13 17 May 2017, 05:23
3 A division of a company consists of seven men and five women 6 05 Jul 2015, 03:47
5 If there are twice as many women as men in a group and an eq 8 15 Nov 2015, 12:31
27 Seven men and five women have to sit around a circular table 30 10 Jan 2017, 08:42
24 Seven men and seven women have to sit around a circular 12 12 Jul 2016, 07:00
Display posts from previous: Sort by