Probability question about married couples

Let us first the label the couples as $A1,A2,A3$ and $A4$.Define the events $E1,E2,E3,E4$ as:

E1 = Event of A1 sitting together.
E2 = Event of A2 sitting together.
E3 = Event of A3 sitting together.
E4 = Event of A4 sitting together.

Hence,the probability $P\space(E1 \cup E2 \cup E3 \cup E4)$ will give the probability of the case where at least one of the four couples will sit together.

So, your required answer is $1-\space$$P\space(E1 \cup E2 \cup E3 \cup E4)$

PS:We can use the principle of mutual inclusion and exclusion to find $P\space(E1 \cup E2 \cup E3 \cup E4)$ and it should give $\frac{23}{35}$ (if I haven't made any mistake with calculation).

Assuming you seat the 8 individuals at random, one way of doing this is to use the inclusion exclusion principle and turn a number of couples into virtual individuals so they must sit next to each other to get $$1-\frac{ 2^1{4 \choose 1} 7! - 2^2{4 \choose 2} 6! + 2^3{4 \choose 3} 5! - 2^4{4 \choose 4} 4!}{8!}$$