How many ways to visit $6$ cities twice?

Question

A person wishes to visit $6$ cities, each exactly twice, and never visiting the same city twice in a row. In how many ways can this be done ?

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I tried by inclusion exclusion.

But Having a problem in finding the total number of outcomes ?

I guess total number of outcomes can be found by using multinomial coefficients like $(12!)/(2!)(2!)(2!)(2!)(2!)(2!)$.

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Am i proceeding correct ?

Answer 1

Proceeding via inclusion-exclusion: There are $2^{n-6}(12-n)!$ arrangements in which $n$ particular cities are visited twice in a row, so there are $$ 2^{-6}\sum_{n=0}^6(-2)^n\binom6n(12-n)!=2631600 $$ arrangements in which no city is visited twice in a row.