TITLE: Find the number of ways that 5 boys and 6 girls can stand in a row if no boy stands next to another boy
QUESTION [6 upvotes]: The question is:

Find the number of ways that 5 boys and 6 girls can stand in a row if no boy stands next to another boy

In which the correct model would be:
$$\square g \square g \square g \square g \square g \square g\square$$
Therefore:
$$6!\cdot 7P5=1\space 814\space 400$$
However, I noticed that if you take one of the girls away (let's label her as $^a$), we have:
$$^a\square^ag^a\square^ag^a\square^ag^a\square^ag^a\square^ag^a\square^a$$
We see that no matter where girl$^a$ stands, the boys will still not stand next to each others:
$$5!\cdot6P5\cdot12=1\space036\space800$$
Update
Also, if we are to swap it around:
$$\square b \square b \square b \square b \square b \square$$
We now have:
$$5!\cdot6!=86\space 400 $$
What is wrong with the last two logic?
Thank you.

REPLY [4 votes]: Your first computation is correct.
The latter two computations undercount because they start off by taking one of the girls away. Let's say they always take girl #1 away.
Now if we number the girls, those solutions can contain as an example $1 2 b 3 b 4 b 5 b 6b$. But they can not contain $1b2b34b5b6b$. We've lost the ability to have two girls next to eachother without involving girl #1.