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In a room are five chairs to accommodate 3 people, one

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In a room are five chairs to accommodate 3 people, one  [#permalink]

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New post 12 Mar 2014, 02:46
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In a room are five chairs to accommodate 3 people, one person to a chair. How many seating arrangements are possible?

A. 45
B. 60
C. 72
D. 90
E. 120

I found this problem tricky.

If you like this problem, please, be generous to give a Kudo.
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Posts: 49251
Re: In a room are five chairs to accommodate 3 people, one  [#permalink]

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New post 12 Mar 2014, 04:12
aja1991 wrote:
In a room are five chairs to accommodate 3 people, one person to a chair. How many seating arrangements are possible?

A. 45
B. 60
C. 72
D. 90
E. 120

I found this problem tricky.

If you like this problem, please, be generous to give a Kudo.


\(C^3_5*3!=60\), where \(C^3_5\) is the number of ways to choose 3 chairs for 3 people, and 3! is the number of arrangements of 3 people on those chairs.

Or directly: \(P^3_5=60\): choosing 3 out of 5, where order matters.

Answer: B.
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Re: In a room are five chairs to accommodate 3 people, one  [#permalink]

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New post 12 Mar 2014, 15:16
I actually solved this problem by the following thoughts (derived from factorials):
First Person: 5 Chairs to choose
Second Person: 4 Chairs to choose
Third Person: 3 Chairs to choose
Then I simply multiplied this values and got the answer. Basically it’s 5! – 2!.

Now the interesting part: I just did that because I had no other idea, how to tackle this problem. Would someone please tell me, if I was just lucky? (and probably explain it differently)
Unfortunately I can’t understand what you are exactly calculating in your solution Bunuel.
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Re: In a room are five chairs to accommodate 3 people, one  [#permalink]

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New post 13 Mar 2014, 00:29
1
holdem wrote:
I actually solved this problem by the following thoughts (derived from factorials):
First Person: 5 Chairs to choose
Second Person: 4 Chairs to choose
Third Person: 3 Chairs to choose
Then I simply multiplied this values and got the answer. Basically it’s 5! – 2!.

Now the interesting part: I just did that because I had no other idea, how to tackle this problem. Would someone please tell me, if I was just lucky? (and probably explain it differently)
Unfortunately I can’t understand what you are exactly calculating in your solution Bunuel.


You used the Basic Counting Principle which is absolutely fine.
Answer is 5*4*3 = 60 (which by the way, is not the same as 5! - 2! = 118)

Here are some discussions on Basic Counting Principle and more:
http://www.veritasprep.com/blog/2011/10 ... inatorics/
http://www.veritasprep.com/blog/2011/10 ... ts-part-i/
http://www.veritasprep.com/blog/2011/10 ... s-part-ii/
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Re: In a room are five chairs to accommodate 3 people, one  [#permalink]

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New post 25 Nov 2017, 02:56
Bunuel wrote:
aja1991 wrote:
In a room are five chairs to accommodate 3 people, one person to a chair. How many seating arrangements are possible?

A. 45
B. 60
C. 72
D. 90
E. 120

I found this problem tricky.

If you like this problem, please, be generous to give a Kudo.


\(C^3_5*3!=60\), where \(C^3_5\) is the number of ways to choose 3 chairs for 3 people, and 3! is the number of arrangements of 3 people on those chairs.

Or directly: \(P^3_5=60\): choosing 3 out of 5, where order matters.

Answer: B.



-------------------------------------------------------------------------------------------------------------------------------------------------------------
Why the two empty chairs are not accounted for? I mean why it's not \(C^3_5*3!*2!\) , where 2! is the empty chairs. Because the three person can have 2 adjacent empty chairs in between them or no empty chair or alternate empty chairs.
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Re: In a room are five chairs to accommodate 3 people, one  [#permalink]

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New post 25 Nov 2017, 04:14
GMATMBA5 wrote:
Bunuel wrote:
aja1991 wrote:
In a room are five chairs to accommodate 3 people, one person to a chair. How many seating arrangements are possible?

A. 45
B. 60
C. 72
D. 90
E. 120

I found this problem tricky.

If you like this problem, please, be generous to give a Kudo.


\(C^3_5*3!=60\), where \(C^3_5\) is the number of ways to choose 3 chairs for 3 people, and 3! is the number of arrangements of 3 people on those chairs.

Or directly: \(P^3_5=60\): choosing 3 out of 5, where order matters.

Answer: B.



-------------------------------------------------------------------------------------------------------------------------------------------------------------
Why the two empty chairs are not accounted for? I mean why it's not \(C^3_5*3!*2!\) , where 2! is the empty chairs. Because the three person can have 2 adjacent empty chairs in between them or no empty chair or alternate empty chairs.


I don;t understand what you mean. The chairs are standing in a row. We are accommodating 3 people.
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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

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: In a room are five chairs to accommodate 3 people, one  [#permalink]

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New post 25 Nov 2017, 05:47
-------------------------------------------------------------------------------------------------------------------------------------------------------------
Why the two empty chairs are not accounted for? I mean why it's not \(C^3_5*3!*2!\) , where 2! is the empty chairs. Because the three person can have 2 adjacent empty chairs in between them or no empty chair or alternate empty chairs.[/quote]

I don;t understand what you mean. The chairs are standing in a row. We are accommodating 3 people.[/quote]


------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Suppose P1C, P2C and P3C denote three persons sitting on three different chairs and EC denotes an empty chair.

With \(C^3_5*3!\), we determined the pattern of three persons sitting on 5 chairs, with interchanging seats.
But is the seating arrangement P1C-EC-P2C-EC-P3C not different from P1C-P2C-EC-EC-P3C. Do we not have to consider empty chairs as well for forming the pattern or is it that which chairs are left empty doesn't matter.

------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
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Posts: 6787
In a room are five chairs to accommodate 3 people, one  [#permalink]

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New post 25 Nov 2017, 06:00
1
Quote:
Quote:
GMATMBA5 wrote:
-------------------------------------------------------------------------------------------------------------------------------------------------------------
Why the two empty chairs are not accounted for? I mean why it's not \(C^3_5*3!*2!\) , where 2! is the empty chairs. Because the three person can have 2 adjacent empty chairs in between them or no empty chair or alternate empty chairs.


I don;t understand what you mean. The chairs are standing in a row. We are accommodating 3 people.



------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Suppose P1C, P2C and P3C denote three persons sitting on three different chairs and EC denotes an empty chair.

With \(C^3_5*3!\), we determined the pattern of three persons sitting on 5 chairs, with interchanging seats.
But is the seating arrangement P1C-EC-P2C-EC-P3C not different from P1C-P2C-EC-EC-P3C. Do we not have to consider empty chairs as well for forming the pattern or is it that which chairs are left empty doesn't matter.

------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------



Hi...

when you are choosing 3 chairs out of 5 by 5C3, you are already catering for where the E are ..
so FFFEE is taken as a different scenario than FFEEF....
Now these three can be filled in 3! ways that is why 5C3*3!

let three person be A, B and C
here 3!=3*2=6 means FFFEE is
ABCEE
ACBEE
BACEE
BCAEE
CABEE
CBAEE

hope it helps
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In a room are five chairs to accommodate 3 people, one &nbs [#permalink] 25 Nov 2017, 06:00
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