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Someone is planning a round-the-world trip that involves visiting $2n$ cities, with two cities from each of $n$ different countries. He can choose a city to start and end the journey in, with the other $2n-1$ cities being visited exactly once. However, he has the restriction that the two cities from each country should not be visited consecutively. How many different trips are possible?

I have used the inclusion exclusion principle and reached that the desired number should be:

$$(2n)!-2n\sum_{k=1}^{n}(-1)^{k-1}\binom{n}{k}2^k \binom{2n-2-(k-1)}{k-1}(k-1)! (2n-2k)!$$

But I am having trouble of finding some closed formula for the sum. I was thinking of using the formula for the product of exponential generating functions, but I have trouble with the large binomial coefficient.

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Your edited version is correct but unnecessarily complicated. Consider the cities arranged in a circle. You want no two adjacent cities to be in the same country, and you want to distinguish one city as the start and end of the journey. There are $\binom nk$ ways to choose $k$ countries for which the constraint is violated, $2^k$ ways to choose the order within those $k$ pairs, $\frac{(2n-k)!}{2n-k}=(2n-k-1)!$ ways to arrange the $2n-k$ objects ($k$ pairs and $2(n-k)$ unpaired cities) in a circle without distinguishing a starting point, and $2n$ ways to distinguish a starting point, for a total of

$$ 2n\sum_{k=0}^n(-1)^k\binom nk2^k(2n-k-1)!\;. $$

This is the same as your result, since your two factorials in the end cancel the denominator of the binomial coefficient and the term $(2n)!$ is the $k=0$ term.

This is ${}_1F_1(-n;1-2n;-2)$, where ${}_1F_1$ is Kummer's confluent hypergeometric function. Wolfram|Alpha provides an alternative representation in terms of modified Bessel functions:

$$ {}_1F_1(-n;1-2n;-2)=\frac1{2^{n+\frac12}\mathrm e}\Gamma\left(\frac12-n\right)\left(I_{\frac12(-2n+1)}(1)+I_{\frac12(-2n-1)}(1)\right)\;. $$

Anyway, the argument that Henry linked to still applies, so for $n\to\infty$ the count is asymptotic to

$$\frac{(2n)!}{\mathrm e}\;.$$

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Apart from being a count, rather than a probability, this is essentially the same as Showing probability no husband next to wife converges to $e^{-1}$ so you can rewrite your expression as $$\displaystyle\sum_{i=0}^n (-2)^i {n \choose i}(2n-i)!$$ which will be close to $$(2n)!\,e^{-1}$$

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  • $\begingroup$ OEIS A007060 $\endgroup$ – Henry Jun 6 '16 at 15:33
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    $\begingroup$ I think there's a difference in that here the start and end of the journey are in the same city and the other city of that country should be visited neither first, nor last. We're essentially seating the couples at a round table without regarding rotated configurations as equivalent. $\endgroup$ – joriki Jun 7 '16 at 6:37

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