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

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Intern
Joined: 03 Dec 2013
Posts: 16
Location: Uzbekistan
Concentration: Finance, Entrepreneurship
GMAT 1: 620 Q42 V33
GRE 1: Q790 V400
GPA: 3.4
WE: Analyst (Commercial Banking)
In a room are five chairs to accommodate 3 people, one  [#permalink]

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12 Mar 2014, 02:46
2
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Difficulty:

5% (low)

Question Stats:

78% (00:45) correct 22% (01:07) wrong based on 176 sessions

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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.
Math Expert
Joined: 02 Sep 2009
Posts: 49251
Re: In a room are five chairs to accommodate 3 people, one  [#permalink]

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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.

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Joined: 21 Jan 2014
Posts: 13
Location: Germany
Concentration: Entrepreneurship, Strategy
GMAT Date: 05-28-2014
GPA: 3.9
WE: Consulting (Entertainment and Sports)
Re: In a room are five chairs to accommodate 3 people, one  [#permalink]

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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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Joined: 16 Oct 2010
Posts: 8281
Location: Pune, India
Re: In a room are five chairs to accommodate 3 people, one  [#permalink]

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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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Intern
Joined: 07 Aug 2017
Posts: 14
Location: India
GPA: 4
WE: Information Technology (Consulting)
Re: In a room are five chairs to accommodate 3 people, one  [#permalink]

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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.

-------------------------------------------------------------------------------------------------------------------------------------------------------------
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.
Math Expert
Joined: 02 Sep 2009
Posts: 49251
Re: In a room are five chairs to accommodate 3 people, one  [#permalink]

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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.

-------------------------------------------------------------------------------------------------------------------------------------------------------------
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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Posts: 14
Location: India
GPA: 4
WE: Information Technology (Consulting)
Re: In a room are five chairs to accommodate 3 people, one  [#permalink]

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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.

------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Math Expert
Joined: 02 Aug 2009
Posts: 6787
In a room are five chairs to accommodate 3 people, one  [#permalink]

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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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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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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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