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

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A group of four women and three men have tickets for seven a [#permalink] New post 30 Dec 2009, 20:43
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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

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Re: Combinations Problem -- Arrangement of Seats [#permalink] New post 30 Dec 2009, 23:31
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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!
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Re: Combinations Problem -- Arrangement of Seats [#permalink] New post 31 Dec 2009, 00:33
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!
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Re: Combinations Problem -- Arrangement of Seats [#permalink] New post 31 Dec 2009, 02:35
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
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Re: Combinations Problem -- Arrangement of Seats [#permalink] New post 31 Dec 2009, 18:53
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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!.

Answer: C.

Hope it's clear.
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Re: Combinations Problem -- Arrangement of Seats [#permalink] New post 01 Jan 2010, 09:07
C is the answer!

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!
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Re: A group of four women and three men have tickets for seven a [#permalink] New post 21 Sep 2013, 11:40
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Re: Combinations Problem -- Arrangement of Seats [#permalink] New post 21 Sep 2013, 11:46
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!.

Answer: C.

Hope it's clear.


Just wanted to share this little thing Bunuel.

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!
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Re: A group of four women and three men have tickets for seven a [#permalink] New post 29 Dec 2013, 16:18
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!

Hence answer is (C)

Hope it helps
Cheers!
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Re: Combinations Problem -- Arrangement of Seats [#permalink] New post 11 Jun 2014, 20:52
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!.

Answer: C.

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
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Re: Combinations Problem -- Arrangement of Seats [#permalink] New post 12 Jun 2014, 03:53
Expert's post
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!.

Answer: C.

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.
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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Combinations Problem -- Arrangement of Seats   [#permalink] 12 Jun 2014, 03:53
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