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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.

\(C^3_5*3!=60\), 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.

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.

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)

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.

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.
_________________

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 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.

------------------------------------------------------------------------------------------------------------------------------------------------------------- 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.

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