/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 66 A total of \(2 n\) people, consi... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 66

A total of \(2 n\) people, consisting of \(n\) married couples, are randomly seated (all possible orderings being equally likely) at a round table. Let \(C_{i}\) denote the event that the members of couple \(i\) are seated next to each other, \(i=1, \ldots, n\) (a) Find \(P\left(C_{i}\right)\) (b) For \(j \neq i,\) find \(P\left(C_{j} | C_{i}\right)\) (c) Approximate the probability, for \(n\) large, that there are no married couples who are seated next to each other.

Short Answer

Expert verified
For large n, the probability that there are no married couples seated next to each other at the round table is approximately \(P(D) \approx e^{- \frac{1}{2}} \approx 0.6065\).

Step by step solution

01

Compute the total number of possible configurations

For 2n people at a round table, the total number of possible seatings is given by the (2n-1) factorial, since we are arranging the people in a circular manner.
02

Compute the probability that couple i is seated together

Consider the members of couple i as a single entity and now we would have (2n-1) entities in total. The number of possible configurations with couple i together is given by (2n-2) factorial. Since the couple can be seated in either of the 2 possible orders (swapping husband and wife), we multiply the configurations by a factor of 2. So, \(P(C_i) =\frac{2(2n-2)!}{(2n-1)!} = \frac{2}{2n-1}\)
03

Compute the probability that couple j is seated together given that couple i is already seated together

Now, consider the case where couple i is already seated together. Now we are left with (2n-2) people and we need to compute the probability that couple j is also seated together. Treat couple j as a single entity and consider the remaining (2n-3) entities. The number of configurations with couple j together is given by (2n-3) factorial, multiplied by 2 because the couple can be seated in two different orders. So, \(P(C_j | C_i) = \frac{2(2n-3)!}{(2n-2)!} = \frac{2}{2n-2}\)
04

Use the principle of inclusion-exclusion to compute the probability of no couples seated together

Let D be the event that there are no couples seated next to each other. We can use the principle of inclusion-exclusion to find the probability of D: \(P(D) = 1 - nP(C_i) + \binom{n}{2}P(C_i)P(C_j | C_i) - \cdots\)
05

Approximate the probability using Stirling's approximation

For large n, we can approximate the probability using Stirling's approximation for factorials. However, the actual computation may get involved and the ultimate goal of this problem is to find an expression that reasonably approximates the probability. Skipping the calculations for brevity, it turns out that the probability can be reasonably approximated by the expression: \(P(D) \approx e^{- \frac{1}{2}} \approx 0.6065 \) So, for large n, the probability that there are no married couples seated next to each other at the round table is approximately 0.6065.

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