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There are $12$ people including A,B and C. They are arranged in a circle. Find the probability that no two among A, B and C are together.

I have solved problem where cases involving two person sitting adjacent to each other. But here there are $3$ people where no two among them sits adjacent, hence I am getting confused.

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I think it is easier to find the probability where the condition is not satisfied. We have two cases to consider:

Case 1: A,B,C are adjacent. This is simply grouping A,B,C and changing their places among themselves with $3!$ and merging them with the remaining $9$ people. In this case, there are

$$3!\cdot(10-1)! = 3!\cdot9!$$

such arrangements.

Case 2: A,B are adjacent, C is not adjacent to neither A nor B. We can find this by grouping $4$ people. While grouping them, put A and B on the two middle places with $2!$ and for the side places, we can choose $2$ people from $9$ (Not $10$ because we don't want C to be one of these side places as it is already counted in case 1) and change their places among themselves with $\binom{9}{2}2! = 72$. Then we can take this group as one person and merge them with the remaining $8$ people. In this case, there are

$$2!\cdot 72\cdot(9-1)! = 72\cdot 2! \cdot 8! = 16\cdot9!$$

such arrangements. Notice that this is the same for the cases where A,C are adjacent B is non-adjacent to neither A nor C; and B,C are adjacent A is non-adjacent to neither B nor C. So, in total, we have

$$6\cdot9!+3\cdot16\cdot9! = 54\cdot9!$$

arrangements that do not satisfy the condition. So the answer should be

$$\frac{(12-1)!-54\cdot9!}{(12-1)!} = \frac{11\cdot10\cdot9!-54\cdot9!}{11!} = \frac{56\cdot9!}{11!} = \frac{28}{55}$$

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    $\begingroup$ In Case 2, you made a minor arithmetic mistake. Notice that $72 \cdot 2! \cdot 8! = 9 \cdot 8 \cdot 2 \cdot 8! = 8 \cdot 2 \cdot 9! = \color{red}{16} \cdot 9!$. If you then correct the subsequent calculations, your answer will agree with mine. $\endgroup$ – N. F. Taussig May 26 at 20:10
  • $\begingroup$ Ah, you are right, thank you very much for the correction. $\endgroup$ – ArsenBerk May 26 at 20:22
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We will seat $A$, $B$, and $C$ after we have seated the others.

Suppose one of the other people is person $D$. Seat her somewhere. It does not matter where since only the relative order of the people matters in a circular permutation. As we proceed clockwise around the the table from person $D$, the other eight people may be seated in $8!$ ways. This creates nine spaces in which to seat persons $A$, $B$, and $C$, one to the left of each of the nine people already at the table. To ensure that no two of $A$, $B$, and $C$ are adjacent, we choose three of these nine spaces. Relative to person $D$, $A$, $B$, and $C$ can be arranged in these spaces in $3!$ ways. Hence, the number of seating arrangements of the $12$ people at a round table in which no two of $A$, $B$, and $C$ are adjacent is $$8!\binom{9}{3}3!$$ Relative to person $D$, the other eleven people may be seated in $11!$ orders as we proceed clockwise around the table. Hence, the probability that no two of $A$, $B$, and $C$ are seated together is $$\frac{8!\binom{9}{3}3!}{11!}$$

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