A Combinatorics Problem: Where Did I Go Wrong?

Question

I am trying to solve a combinatorics problem, specifically problem 2iii). I have an answer, but the textbook says it's wrong:

What I did was 12! - 11! - 10! - 9! - 8!, because 12! is the total number of combination, and 11! is the number of combinations if we have 2 girls together, etc.

I would appreciate it if you would do 2 things:

1. Can you explain to me where I went wrong?
2. Also, can you explain to me why the textbook's answer is $7!\binom855!$

Thanks in advance!

Answer 1

Edit after OP's edit:

Your textbook is right concerning (iii).

Let's analyze the situation:

• For the $7$ boys you get: $\Rightarrow \color{blue}{7!}$
• When you line the boys up, they provide $8$ "slots" where to put exactly one girl. So, choose $5$ of these $8$ slots: $\Rightarrow \color{blue}{\binom 8 5}$
• The $5$ girls themselves can also be arranged: $\Rightarrow \color{blue}{5!}$

All together $$\color{blue}{7! \cdot \binom 8 5 \cdot 5!}$$

Note that I cannot explain where you made a mistake as you need to give more details about your reasoning.

Answer 2

Fix AGGGB as one then this plus seven remaining can be seated in 8! ways. The three girls between A and B can be chosen in ${5\choose3}$ ways and the three girls can be permuted in 3! ways to a total of $8!{5\choose3}3!$. The two boys A and B can be permuted in 2! ways to a total of $8!{5\choose3}3!2!$