No husband can sit next to his wife in this probability question I have a probability question that reads:
Question: 
If 4 married couples are arranged in a row, find the probability that no husband sits next to his wife.
My attempt:
Total outcomes = 8!
Outcomes that all of them sit with their wife: 4!*(4*2!)
Outcomes that one of them sit with their wife: (2!*4)(6!)-(2!*4)(3!*(3*2!)[subtract the ways that remaining couples are together]
Outcomes that two of them sit with their wife: (2!*4C2)(4!)-(2!*4C2)(2!*(2*2!)
Outcomes that three of them sit with their wife: (2!*4C3)(2!)-(2!*4C3)(1!*(2*2!)
Hence no husband sit with wife is 1 -(Outcomes that all of them sit with their wife+ Outcomes that one of them sit with their wife + Outcomes that two of them sit with their wife + Outcomes that three of them sit with their wife)/8!

Am i right? Any easier way?
 A: I cannot think of anything pleasant. A natural approach is through Inclusion/Exclusion. 
There are $8!$ arrangements. If we can count the bad arrangements, in which at least one couple is together, then the rest is easy.
Call the couples A, B, C, D and let $X$ be the wife in couple X, and $x$ the husband. It is not hard to count the arrangements in which $a$ is next to $A$, and similarly for the other $3$ couples.
If we add these $4$ numbers, we will have double-counted, in particular, the arrangements in which couple A and couple B are both together. So we need to subtract $\binom{4}{2}$ times the number of arrangements in which couple A and couple B are together.
But we have subtracted too much. So we add back $\binom{4}{1}$ times the number of arrangements in which couples A, B, and C are together.
But we have added back too much, so we must subtract the number of ways for all the couples to be next to each other. 
Instead of counting, we can apply Inclusion/Exclusion directly to probabilities. It is marginally easier.
A: With $n$ couples, the probability of no couple sitting together is $$\displaystyle\sum_{i=0}^n (-2)^i {n \choose i}\frac{(2n-i)!}{(2n)!}.$$ 
This comes from using inclusion-exclusion, and (as I said in the previous question) treating a couple sitting together as a single individual (hence the $(2n-1)!$), though one of each pair can be on the left or right (hence the $2^i$).  
