Probability question about married couples 
If four married couples are arranged in a row, what is the probability 
  that no husband sits next to his wife?

Would it be
$1- \frac{2(4!)}{8!}$? 
 A: Let's count the complement. If at least one couple stands together, you have $\binom{4}{1} = 4$ ways of choosing the couple, $2$ ways of arranging them, and $7$ positions in which to place them (the first person in the couple cannot stand in the last place of the line), and the remaining 6 positions can be filled in $6!$ ways; this gives a total of $4\times 2\times 7\times 6! = 8!=40320$ ways. 
However, this overcounts by counting arrangements in which more than one couple sits together twice. If at least two couples stand together, we have $4\times 3$ ways of choosing the first and second couple, $4$ ways of arranging them, 15 ways to stand them in the line in that order, and $4!$ of filling in the remaining 4 places, for a total of $4\times 3\times 4\times 15\times 4! = 17280$ ways. 
If at least 3 couples stand together, we have $4\times 3\times 2$ ways of choosing the couples (in order), $8$ ways of arranging them amongst themselves, 10 ways of arranging them in the line, and $2$ ways of filling the remaining seats, for a total of $4\times 3\times 2\times 8\times 10\times 2= 3840$ ways.
If all four couples sit together, we have $4\times 3\times 2\times 1$ way of arranging the couples, and $2^4 = 16$ ways of arranging each couple, for a total of $384$ ways.
So, let's see: when we count "at least one couple stands together", we count the arrangements with exactly one couple together once, the arrangements with exactly two couples together twice; the arrangements with exactly three couples three times, and the arrangements with all four couples together four times.
When we could "at least two couples together", we count the arrangements with exactly two couples together once, the arrangements with exactly three couples together $\binom{3}{2}=3$ times; and the arrangements with all four couples together $\binom{4}{2}=6$ times.
When we could "at least three couples together", we count the arrangements with exactly three couples together once, and the arrangements with all four couples together $\binom{4}{3}=4$ times.
So if we take:
$$(\text{at least one}) - (\text{at least two}) + (\text{at least three}) - (\text{all four})$$
we get the count exactly right.
A: 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 $$1-\frac{ 2^1{4 \choose 1} 7! - 2^2{4 \choose 2} 6! + 2^3{4 \choose 3}  5! - 2^4{4 \choose 4} 4!}{8!}$$ 
A: 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).
