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I have purchased a book to learn combinatorics and probability, but it does not come with solutions to its exercises. I would like to know if I am on the right track, so any help with this problem would be appreciated. Thank you.

1. There is a round table with $8$ identical chairs around it. Prem, Victor and 6 other people will be randomly assigned around it.

(a) How many arrangements are there such that Prem and Victor are sitting next to each other?

Thinking of them as a single person gives us $7$ people to seat. This means a $7!$ combination. We can split the ways they can be seated in $2$ and must divide $7$ to avoid double counting. $$\frac{7!}{7} \cdot 2 = 6! \cdot 2$$

(b) What is the probability that Prem and Victor will sit next to each other? What is the probability that they don’t sit next to each other? Which is more likely?

All possible ways minus the chances they sit next to each other versus the chances they do sit next to each other.

chances that they do not sit together: $$\frac{8!}{8} – 6! \cdot 2 = \frac{3600}{5040}$$ chances that they do sit together: $$6! \cdot 2 = \frac{1440}{5040}$$ More likely they do not sit together.

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How many arrangements of eight people sitting at a round table with identical chairs are there in which Prem and Victor sit next to each other?

You are correct. Seat Prem. Victor can sit to his left or right. Once they are seated, the remaining people can be seated in $6!$ ways as we proceed clockwise around the circle from the block of seats occupied by Prem and Victor. Therefore, there are indeed $$2 \cdot 6!$$ ways to seat the eight people if Prem and Victor sit in adjacent seats.

What is the probability that Prem and Victor will sit next to each other? What is the probability that they don't sit next to each other? Which is more likely?

Once we seat Prem, the remaining seven people can be seated in $7!$ ways as we proceed clockwise around the circle from Prem. Since there are $2 \cdot 6!$ seating arrangements in which Prem and Victor are adjacent, the probability that they are seated next to each other is $$\frac{2 \cdot 6!}{7!} = \frac{2}{7}$$ While your answer $1440/5040$ reduces to $2/7$, the equation you wrote is false since $$\frac{8!}{8} - 6! \cdot 2 \neq \frac{3600}{5040}$$ You forgot to include the denominator $7!$ on the left-hand side. You should have written $$\frac{\frac{8!}{8} - 6! \cdot 2!}{7!} = \frac{3600}{5040}$$

The probability that they do not sit together is found by subtracting the probability that they do sit together from $1$, so we obtain $$1 - \frac{2}{7} = \frac{5}{7}$$ Your answer $3600/5040$ is equivalent to this, but the equation you wrote is false. Again, you forgot to include the denominator of $7!$ on the left hand side.

As you observed, it is indeed more likely that Prem and Victor do not sit next to each other if they are seated randomly.

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