Let us count the number of quadrilaterals $A_1A_2A_{2+i}A_{2+i+j}$ where $i >1, j > 1, 2+i+j < 12$. Such vertices count the quadrilaterals with exactly $A_1A_2$ as common side. This is same as the number of solutions to $i+j <10, i >1, j > 1 $. Putting $x = i-1, y = j-1$, we need the solutions $x+y < 8$ where $x >0, y > 0$. By stars and bars method this is $\binom{1}{1}+\binom{2}{1}+\binom{3}{1}+\binom{4}{1}+\binom{5}{1} +\binom{6}{1}= 21$. Thus the answer for part 1 is $21\times 12=252$.
Here's an ad hoc solution that doesn't generalize well to higher numbers of balls.
I'll treat balls of the same colour as distinguishable throughout and divide through by the number $4!3!2!2!$ of permutations of indistinguishable balls in the end.
Since it's the $4$ green balls that make inclusion-exclusion most cumbersome to apply, let's regard these as walls and distribute the remaining balls among them. If we have $k$ non-green objects, we can select one for each green ball in $\frac{k!}{(k-4)!}$ ways, glue the green balls to their right, and then permute the resulting $k$ objects in $k!$ ways. This doesn't allow a green ball to be on the very left, so we have to add a contribution where we select one of the $4$ green balls to be on the very left, select one object for each of the remaining $3$ green balls in $\frac{k!}{(k-3)!}$ ways and permute the resulting $k$ objects in $k!$ ways. Let's denote the total for $k$ objects by $g_k$:
$$
g_k=k!^2\left(\frac1{(k-4)!}+\frac4{(k-3)!}\right)\;,
$$
where terms with negative factorials should be omitted.
Now we have $2$ red, $2$ yellow and $3$ blue balls to distribute. That yields $\binom22+\binom22+\binom32=5$ adjacency conditions to satisfy. Let's first ignore the complications from the interactions of the blue pairs and just perform inclusion-exclusion for these five conditions. This yields
$$
\sum_{k=0}^5(-1)^k\binom5ka_k\;,
$$
where $a_k=2^kg_{7-k}$ is the number of arrangements that result when we glue $k$ pairs of the $7$ non-green balls together and the factor $2^k$ arises because each pair can be glued together in two different orders.
Now we have to account for the interactions of the blue pairs. First consider the cases including two blue adjacency conditions. These are in fact satisfiable, but we overcounted them, since there are only $2$ orders in which the blue balls can be glued together to satisfy two adjacency conditions, whereas we counted $4$, so we have to substract half of that contribution:
$$
\frac12\binom32\sum_{k=0}^2(-1)^{k+2}\binom2ka_{k+2}\;,
$$
where $k$ now counts the non-blue adjacency conditions fulfilled.
That leaves the cases with all three blue pairs. These shouldn't have been counted at all, since it's impossible to have all three pairs of blue balls adjacent, so we need another subtraction of
$$
\sum_{k=0}^2(-1)^{k+3}\binom2ka_{k+3}\;.
$$
Adding it all up yields
\begin{align}
&a_0-5a_1+10a_2-10a_3+5a_4-a_5-\frac32\left(a_2-2a_3+a_4\right)+(a_3-2a_4+a_5)\\
={}&a_0-5a_1+\frac{17}2a_2-6a_3+\frac32a_4\\
={}&8467200-5\cdot1209600+\frac{17}2\cdot172800-6\cdot23040+\frac32\cdot2304\\
={}&3753216\;,
\end{align}
and dividing by $4!3!2!2!=576$ yields $6516$ admissible arrangements, in agreement with the calculation on the site you linked to.
Best Answer
We will solve the problem for a row first, then subtract those arrangements in which two people will be placed in adjacent seats if we join the ends of the row together to form a circle.
We will arrange nine blue and six green balls. Line up the nine blue balls. This creates ten spaces in which we could place a green ball, eight between successive blue balls and two at the ends of the row.
To separate the green balls, we choose six of these ten spaces in which to insert a green ball, which can be done in $\binom{10}{6}$ ways. For instance, if we choose the first, third, fifth, sixth, eighth, and ninth spaces, we obtain the arrangement shown below.
We now number the balls from left to right. The numbers on the green balls represent the selected seats. Thus, if the seats were arranged in a row, there would be $$\binom{10}{6}$$ ways to select the seats so that no two of them were adjacent.
However, if we had an arrangement such as
when we joined the ends of the row, two of the green balls would be adjacent. We must remove these arrangements. Notice that in selecting arrangements with green balls at both ends of the line, we must select both spaces at the ends of the line and four of the eight spaces between successive blue balls in which to place a green ball. Hence, there are $$\binom{8}{4}$$ such arrangements.
Subtracting these from the total yields $$\binom{10}{6} - \binom{8}{4}$$ selections in which no two of the green balls (seat positions) are adjacent when the balls (seats) are arranged in a circle.