It's my second try to solve the question I posted here Combinatorics question - How many different ways to change sitting order
I got some really good advice, but no one said the answer, I solved it, and before I rush into the exam, I'd like to have it verified. I solved it using inclusion exclusion.
Six children (a through f) are playing on a carousel with 6 seats (numbered one to six) such that a is sitting in front of d, b is sitting in front of e, and c is sitting in front of f.
How many ways are there to change the sitting order, such that no child is sitting in front of the child they are sitting in front of now?
First, let us count the total number of combinations. There are 6 seats and 6 children, therefore there are $6!$ different combinations (note that it's $6!$ and not $5!$ since the seats are numbered, and rotating right or left is a totally different order).
let $A$ be the set of sitting arrangements such that $a$ sits infront of $d$.
let $B$ be the set of sitting arrangements such that $b$ sits infront of $e$.
let $C$ be the set of sitting arrangements such that $c$ sits infront of $f$.
Let's count $|A|$. First we need to choose a seat for child $a$. he has $6$ options to choose from. As soon as he chose, child $d$ has no option but to sit infront of him. and now we are left with $4$ seats and $4$ children, there are $4!$ options for this, so $|A|=6*4! = 144$.
Due to symmetry, $|B|=|C|=144$ as well.
let's count $|A\cap B|$. First we choose a seat for $a$, that's $6$ options, $d$ sits infront him of immediatly. then choose a place for $b$, we have $4$ options, $e$ sits infront of him immediately. then we are left with $2$ seats and $2$ children, that's $2!$ options, so overall $|A\cap B|=6*4*2!=48$.
Due to symmetry, $|A\cap C|=|B\cap C|=48$ as well.
it is clear to see that $|A\cap B\cap C|=1$.
$|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C|=3*144-3*48+1=289$.
There are $289$ bad arrangements, and there are $6!=720$ arrangements in total, so there are $720-289=431$ good arrangements and that is my final answer.
Is this correct?