8 people A,B,C,D...H are sitting around an octagonal table.A does not want to sit beside D or opposite to him.B and C wants to sit together. In how many ways can this be done? The answer says (8*4*4*2!*4!)/8. But how?
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asked Nov 14, 2013 at 14:14
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1 Answer
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First we seat $A$, who has $8$ choices. Then there are $4$ places for $D$. There are now $4$ places remaining for $BC$, who can be ordered in $2!$ ways. Now I claim the last factor in the numerator should be $4!$ for the ways to place $EFGH$. We divide by $8$ because the seats are not labeled, so rotating the whole configuration is not counted as changing things.
answered Nov 14, 2013 at 14:22
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$\begingroup$ Thanks Ross, I could solve and understand this problem now. $\endgroup$ Nov 14, 2013 at 14:28
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1$\begingroup$ The secret to these is careful counting. For example, it isn't obvious that the pair BC has the same number of places available after A,D are seated. Given the way the answer was presented it looked likely. I had to sketch it out to be sure of that. Otherwise, you would have to break it into cases depending on where D sits. $\endgroup$ Nov 14, 2013 at 15:08