I've generalized the question I was given here for simplicity: $6$ men and $6$ women are to be paired for a bus trip. If the pairings are done randomly, what's the probability that no women will end up sitting next to a man? Here's my first attempt, but I'm really not sure whether this is the right way to find the desired probability.
We want to group $12$ people into $6$ single-sex groups of $2$, so I began by calculating the number of ways we can make $6$ pairs. By the multinomial function: $12!/(2!)^6=7484400$ ways to make $6$ pairs.
Then, we want to figure out the number of ways we can make single-sex pairs. Again, by the multinomial function, we have $6!/(2!)^3=90$ ways to make $3$ female-female pairs. Since we also have to consider male pairs, I squared this to get $8100$ ways to make $6$ single-sex pairs.
I found that the likelihood of all pairs being single-sex is $8100/7484400=0.00108$, but this doesn't seem like a completely reasonable probability.
Could you help me find the errors in my method for solving this problem?