What is the probability that no two married members sit next to each other? A dancing club consists of 24 people (12 male and 12 female); out of the 24 people,
there are five married couples.
If the members of the club sit in a row randomly, what is the probability that no two married members sit next to each other?
They gave an answer of 0.4960
I'm struggling to get that answer.
I thought of letting the people who are not married be ordered and then place the married people in between them so that no 2 married members sit next to each other.
so there is 7 male and 7 female people who are not married
$\frac{{14!}{15\choose10}{10!}}{24!}$ but it's not right?
Do you think they meant to say people who are married to each other?
 A: So, try sitting down all the single people, and all the married males. There are 19! arrangements. Now one by one add the married females in the gaps (including the ends) The first has a choice of 18 (20 places, two of which are next to her husband), the next 19, up to 22.
This gives:
$$\frac{19!\frac{22!}{17!}}{24!}=\frac{18\cdot19}{23\cdot24}=0.6196$$
Not sure why this isn't the same as your answer.
A: I would approach this problem using the principle of inclusion and exclusion.  Define the events $A, B, C, D, E$ that each of the respective married couples do sit next to each other.  Then counting the ways that no married couple sits together would amount to counting $(A\cup B\cup C\cup D\cup E)^c$.
Then for instance, the number of seatings where $A$ occurs would be $2\cdot 23!$ (Pick an order for couple A (two ways) and then order 22 individuals and the $A$ block ($23!$) ways).
Then also for instance, the number of ways where say $A$ and $B$ both occur would be $2^2\cdot 22!$.  And so on.
Using the Principle of Inclusion and Exclusion (and using $n(S)$ the number of elements in a set $S$), we get that the number of ways for $(A\cup B\cup C\cup D\cup E)^c$ to occur is 
$$24! - n(A)-n(B)-\cdots-n(E)+n(A\cap B) + ...+n(D\cap E)-(\text{triple intersections})...$$
Putting this all together gives that the number of desirable seatings is:
$$24!-5\cdot 2\cdot 23! + {5 \choose 2}\cdot 2^2\cdot 22!-{5\choose 3}\cdot 2^3\cdot 21!+{5\choose 4}\cdot 2^4\cdot 20!-2^5\cdot 19!$$
Dividing the above by $24!$ gave that the probability of a desirable seating is about .6495.  (Different than your answer, so I don't know if I missed something , or if your answer is incorrect.)
