In a room are five chairs to accommodate 3 people. One person to a cha

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

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)

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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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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]


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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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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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