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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]

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New post 30 Dec 2009, 20:43
3
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A
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C
D
E

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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!

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

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New post 31 Dec 2009, 18:53
10
8
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]

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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]

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New post 31 Dec 2009, 00:33
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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]

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New post 31 Dec 2009, 02:35
1
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]

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

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New post 21 Sep 2013, 11:46
1
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]

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New post 29 Dec 2013, 16:18
1
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]

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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]

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New post 12 Jun 2014, 03:53
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!.

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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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

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

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New post 10 Feb 2016, 18:17
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...
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Re: A group of four women and three men have tickets for seven a  [#permalink]

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New post 08 Aug 2018, 20:06
Bunuel
When do we divide the 5!/4 and 3!/3 ... I mean in this problem when we say WWWW(MMM) ...shldnt we divide it by 4? As WWWW are identical .. please respond

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Re: A group of four women and three men have tickets for seven a &nbs [#permalink] 08 Aug 2018, 20:06
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