Selecting 3 persons sitting round a table

[verty]

4 copies? Do you mean 3?

Only 2 should be counted, is what I meant. When n=6, we have ACE or BDF, even if C was selected.

 

[haruspex]

Guys, this problem really isn't that hard.
As I said in post #18, consider each person as taking two adjacent seats, keeping a vacant seat to the left, say. The first pair can go in n positions. That leaves a line of n-2 seats, numbered 1 to n-2 from the left, say. In how many places can the leftmost of the remaining two pairs go, remembering to leave space for the last pair? If they take seats 1 and 2, how many choices for the last pair? If they take seats r and r+1, how many choices for the last pair?

 

[Dick]

Guys, this problem really isn't that hard.
As I said in post #18, consider each person as taking two adjacent seats, keeping a vacant seat to the left, say. The first pair can go in n positions. That leaves a line of n-2 seats, numbered 1 to n-2 from the left, say. In how many places can the leftmost of the remaining two pairs go, remembering to leave space for the last pair? If they take seats 1 and 2, how many choices for the last pair? If they take seats r and r+1, how many choices for the last pair?

Pretty much what I've been saying from the beginning. The direct way actually easy if you just break into two cases depending on where the second person is seated.

 

[Yukoel]

If you select 3, you are done. The seating order of the members is fixed. You can select 1 and treat (3,4,3) and (3,3,4) as different cases afterwards. Hmm... that does work without case-by-case analysis, but then you have to evaluate a sum over n summands.

Well can't we just use (n-3-1)C(3-1) ?(The number of distributing n things in r groups with each of the being nonempty is (n-1)C(r-1) )Here n=n-3 and r=3.

When n = 6, we have only 111 with 4 copies. I mean, we have 6 ways to select the first seat, but 4 are copies with this pattern. When n = 7, we have 112 with 0 copies. When n = 8, 113 and 122 with 0 copies.

When 3 divides n - 3, we get a pattern with copies, 111 or 222 when n = 9. So you have to subtract these I think.

Maybe I misunderstand the problem but the people are treated as distinct right?
So the copies of the same would be for n=6
ACE;CEA;and EAC(The counting is done clockwise only and so is the choosing of gaps).Hence there are three copies of the same thing right? Well yes one can subtract the overcounted things or maybe divide it as well by 3.What I am saying is to fix the first member on the circle and proceed anticlockwise through it with gaps.I do not understand the case with n=7 'cause each single choice has 3 copies.(Something like ACE again)
For n=9 111 doesn't fit on the table so it isn't equivalent to 222 which does fit.Is this wrong ?
Please do tell.
Thanks
Yukoel

 

[verty]

Maybe I misunderstand the problem but the people are treated as distinct right?

Too much thinking for me, sorry. I don't like how the sum depends on n. Let's leave it.

Using cases is definitely quickest. I guess this is a general principle - try cases first.

 

[Saitama]

I was confused by so many replies so I did not reply for sometime but I did figure it out. It will be difficult for me to put it in words but I will try my best.

Selecting three persons without restriction, nC3.
When three persons are sitting together, there are n ways.
In the last case, when two persons are sitting together, there can be n ways to select two persons sitting next to each other and the last person could sit at any of (n-4) places as he cannot sit near any of the two persons.

Hence, total ways=nC3-n-n*(n-4). Solving, I do end up with a right answer.

Thank you everyone!