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In a room are five chairs to accommodate 3 people, one [#permalink]
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12 Mar 2014, 01:46
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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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Re: In a room are five chairs to accommodate 3 people, one [#permalink]
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12 Mar 2014, 03:12



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Re: In a room are five chairs to accommodate 3 people, one [#permalink]
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12 Mar 2014, 14:16
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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12 Mar 2014, 23:29
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 ... tsparti/http://www.veritasprep.com/blog/2011/10 ... spartii/
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Re: In a room are five chairs to accommodate 3 people, one [#permalink]
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25 Nov 2017, 01:56
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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25 Nov 2017, 03:14
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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Re: In a room are five chairs to accommodate 3 people, one [#permalink]
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25 Nov 2017, 04:47
 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 P1CECP2CECP3C not different from P1CP2CECECP3C. 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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In a room are five chairs to accommodate 3 people, one [#permalink]
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25 Nov 2017, 05:00
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 P1CECP2CECP3C not different from P1CP2CECECP3C. 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
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