What is the probability that 5 boys and 5 girls line up such that neither gender is in an uninterrupted block? Just for reference, the exact question is: "Ten children (five boys and five girls) are standing in line. Assume that all possible ways in which they might line up are equally likely. What is the probability that neither the boys nor the girls stand together in an uninterrupted block?"
In general, the number of ways that $10$ children can be lined up is $10!$ 
Let's say there are $5$ boys and $5$ girls.
The number of ways the line could have an uninterrupted block of $5$ girls (or $5$ boys) somewhere in the line is $6 * 5! * 5!$
The number of ways the line could have two uninterrupted blocks is $2  * 5! * 5!$, since the line could either start with the boy block or the girl block 
So the probability that the line doesn't have an uninterrupted block of children should be  $({10! - (6 * 5! * 5!) - (6 * 5!*5!) - (2*5!*5!)})\ / \ 10!$
But that is apparently not the right answer. Instead, the right answer is $(10! - (6 * 5! * 5!) - (6 * 5!*5!) + (2*5!*5!))\ /\ 10!$
I can't imagine why you add the $2*5!*5!$ term instead of subtract. Anyone have any ideas? 
 A: When you do $6 \times 5! \times 5!$ for the block of $5$ boys, $2$ out of the $6$ ways of inserting the boy-block would also result in the girls all in a block, i.e. when you insert the boy-block in front of all the girls, and when you insert the boy-block behind all the girls.  So when you do $2 \times 6 \times 5! \times 5!$, you have double-counted the cases of each group in its own block.  That's why you need to subtract:
$$10! - (6 \times 5! \times 5! + 6 \times 5! \times 5! - 2 \times 5! \times 5!)$$
Hope that answers your question!  
BTW in this particular case, another way to count is to do $4 \times 5! \times 5!$ which counts the number of ways to insert a boy-block into the girls s.t. the girls are definitely not in a block (i.e. just count the 4 positions between girls).  Then you would need to explicitly add the case of each group in its own block, i.e. 
$$10! - (4 \times 5! \times 5! + 4 \times 5! \times 5! +2  \times 5! \times 5!)$$
As you can see, these number expressions have equal value.
A: While subtracting $6\cdot 5!\cdot 5!$ twice, you subtract twice the arrangements "5B5G", "5G5B". So, you need to add them once at the end. Hence, you need to add $2\cdot 5!\cdot 5!$
Another approach could be:


*

*arrangements with exactly one block of boys: $\color{blue}{4}\cdot 5!\cdot 5!$

*arrangements with exactly one block of girls: $\color{blue}{4}\cdot 5!\cdot 5!$

*arrangements with exactly two block of boys and girls: $\color{blue}{2}\cdot 5!\cdot 5!$
So, you get for the numerator:
$$10! - {4}\cdot 5!\cdot 5! - {4}\cdot 5!\cdot 5! - \color{blue}{2}\cdot 5!\cdot 5!$$
