in how many ways can 20 members sit in a round Boardroom if the chairman must sit between the secretary and the treasure?i tried (20-1)!3!*2 but the answer seems to be alittle exaggerated.help me
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- Pick the chairman place: $20$ choices 2. Pick the secretary and treasure places: $2$ choices 3. Place the $17$ remaining members: $17!$ choices. So I'd say $40\cdot17!$.
– NeedForHelp Feb 03 '17 at 06:13 -
1@NeedForHelp: round tables usually are considered not having identifiable seats, so you only count 1 for the chairman-his place defines the orientation. The rest is fine. – Ross Millikan Feb 03 '17 at 06:17
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@Ross Millikan My first language is not English but I think that if the chairman must sit at a specific seat then the question should make that explicit. – NeedForHelp Feb 03 '17 at 06:22
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1@NeedForHelp: it is not that he must sit at a specific seat, it is that we don't care what seat he sits in, because we consider all rotations of a seating pattern equivalent. That is why we specify a round table. If we paint one of the chairs we break the symmetry and your response is correct. – Ross Millikan Feb 03 '17 at 06:25
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Let the group of chairman, secretary and treasurer be one group to be arranged with the 17 others around the circle.
The chairman can be seated in 1 way: in the centre chair of the 'group'; then the secretary and treasurer can either sit respectively on his L and R or his R and L i.e. 2 ways.
Then we arrange this group with the 17 others around a circle in $(18-1)!=\frac{18!}{18}$ ways.
Hence the total number of seating arrangements is $1*2*17!$
M.Diggerson
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Group the $3$ together and then left are $17$. That means you have a total of $18$ people to make them sit in a circle That gives you $17!$ ways to do this. And now for each of these permutations you have an addition multiplication of $2!$ because in that group of $3$ two of them can switch places.Hence you have
$17!.2!$
Upstart
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