- #36
verty
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mfb said: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.
mfb said:4 copies? Do you mean 3?
haruspex said: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?
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.mfb said: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.
Maybe I misunderstand the problem but the people are treated as distinct right?verty said: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.
Yukoel said:Maybe I misunderstand the problem but the people are treated as distinct right?