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

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.

Answer 1

In the last one you're obviously leaving some out, which could be gotten by swapping some of bg combinations in either sum.

In the second, also, you're not counting certain arrangements, which could be gotten by, again, swapping at least some bg combinations.

The right one is the first.

Answer 2

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.