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$10$ people including $x,y,z $ sit around a circular table. How many ways can they

sit together such that $x,y,z$ are not together

Attempt:: number of arrangement in which $x,y,z$ not sit together $ = $ total arrangenent in a circle $-(x,y,z)$ sit together

$\displaystyle = 9! - \left(7!\times 3!\right)$

but it is not right, could some help me to solve it, thanks

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  • $\begingroup$ your anawer seems right to me $\endgroup$ – Abhash Jha Apr 20 '17 at 15:25
  • $\begingroup$ Still there can be a case such that anticlockwise and clockwise order are taken as not different..in that case circular permutation of n diff things taken r at at a time = $\frac{1}{2r}.\frac{n!}{(n-r)!}$ $\endgroup$ – Abhash Jha Apr 20 '17 at 15:33
  • $\begingroup$ It depends what you mean by "not sit together". Do you mean "no two of $x,y,z$ sit together", or do you mean "$x,y,z$ do not sit together as a threesome"? $\endgroup$ – quasi Apr 20 '17 at 15:58
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Seat $x$ first to fix the orientation of the table, then seat $y$ and $z$ linearly arranged with the remaining $7$ identical chairs so that there is at least $1$ chair between all $3$ people in

$$\binom{7+2-3}{2}\cdot 2!$$

ways.

$2!$ is for the order of $y$ and $z$. The binomial coefficient counts linear arrangements of $2$ identical seats for $y$ and $z$ along with the $7$ remaining identical empty chairs such that there is at least $1$ empty chair between $y$ and $z$ and $1$ at either end (this is why we subtract $3$ chairs in the binomial coefficient).

Then seat the remaining $7$ people in the remaining $7$ chairs in $7!$ ways to give

$$\text{arrangements} = \binom{6}{2}\cdot2!\cdot 7!=151\,200\tag{Answer 1}$$

If seats are distinct then multiply that by the number of ways to initially seat $x$ (i.e. by $10$) giving

$$\text{arrangements} = 10\cdot\binom{6}{2}\cdot 2!\cdot 7!=1\,512\,000\tag{Answer 2}$$

The way you did it allows any $2$ of $x$, $y$ and $z$ to be in adjacent seats. We could modify your method by subtracting the ways of seating any $2$ of those $3$ adjacent then add back the cases where all $3$ are adjacent via inclusion-exclusion to give an answer (for which I won't go into in any further detail but which nonetheless checks out with our first answer above)

$$9!-\binom{3}{2}\cdot 2!\cdot 8! +\binom{3}{3}\cdot 3!\cdot 7!=151\,200$$

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We will seat $x$, $y$, and $z$ after seating the other seven people.

Say Anjali is one of the other seven people. We seat her first. While we ask $x$, $y$, and $z$ to wait, we seat the other six people. As we proceed clockwise around the table relative to Anjali, they can be seated in $6!$ ways.

Seating these seven people creates seven spaces, one to the immediate left of each of the seven people currently seated at the table. To ensure that no two of $x$, $y$, and $z$ sit in consecutive seats, we choose three of these seven spaces in which to insert chairs for $x$, $y$, and $z$. We can arrange $x$, $y$, and $z$ in these chairs in $3!$ ways as we proceed clockwise around the table relative to Anjali.

Hence, the number of seating arrangements of ten people at a circular table in which no two of $x$, $y$, and $z$ sit in adjacent seats is $$6!\binom{7}{3}3!$$

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