Number of ways to seat people around a circular table

Question

I got (i) which is $9!$, but there are no answers for the second question. I stated that $$P(\text{none together})=1-P(\text{3 together})-P(\text{2 together})$$ and got the answer $7/12$. Is this the correct answer? Thanks!

Answer 1

Gemma’s lefthand neighbor can’t be Pasha or Rickey, so there are $7$ ways to pick that student; once that person has been picked there are $6$ ways to pick Pasha’s lefthand neighbor, and then there are $5$ ways to pick Rickey’s lefthand neighbor. Think of each of the named students together with his or her lefthand neighbor as a single unit, and think of each of the remaining $4$ students as a single unit. Then there are $7$ units, so there are $6!$ ways to arrange them in a circle. That gives us a total of $7\cdot6\cdot5\cdot6!=30\cdot7!$ arrangements in which no two of the named students are adjacent. Since there are $9!$ arrangements altogether, the probability of this outcome is

$$\frac{30\cdot7!}{9!}=\frac{30}{72}=\frac5{12}\;.$$

Answer 2

$\underline{Sol. 1}$

Arrange the 7 other people in a circle, which can be done in $6!$ ways.

This creates 7 gaps, so the 3 remaining people can choose their separate gaps in $7\cdot6\cdot5$ ways,

and therefore the probability is given by $\displaystyle\frac{6!\cdot7\cdot6\cdot5}{9!}=\frac{6\cdot5}{8\cdot9}=\frac{5}{12}$.

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$\underline{Sol. 2}$

1) If all 3 are together, there are $3!$ ways to arrange the 3 in order, and then $7!$ ways to arrange the remaining 7 people.

2) If exactly 2 are together, there are $\binom{3}{2}$ ways to choose the two together, $2!$ ways to arrange them in order, $6$ ways to choose the place for the third person, and $7!$ ways to arrange the others.

Therefore the probability of none together is given by

$\displaystyle 1-\frac{3!7!+3\cdot2\cdot6\cdot7!}{9!}=1-\frac{3!+36}{8\cdot9}=1-\frac{7}{12}=\frac{5}{12}$.

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$\underline{Sol. 3}$

First seat Pasha. Let $a$, $b$, and $c$ be the number of seats in the gaps between the 3 people, starting with Pasha and moving counterclockwise.

Then $a+b+c=7$ with $a\ge1, b\ge1, c\ge1$, so there are $\binom{6}{2}$ choices for the sizes of these gaps.

There are $2$ ways to arrange Gemma and Rickey in their seats, and $7!$ ways to arrange the other 7 people; so the probability that none of the 3 sit together is given by

$\displaystyle\frac{\binom{6}{2}\cdot2\cdot7!}{9!}=\frac{15\cdot2}{8\cdot9}=\frac{5}{12}$.