Place the saucers on the table in the order: azure, azure, blue, blue, cyan, cyan.
Let's treat cups of the same color as indistinguishable.
There are $\binom{6}{2}$ ways to choose two of the six positions for the azure cups, $\binom{4}{2}$ ways to choose two of the remaining four positions for the blue cups, and $\binom{2}{2}$ ways to choose both of the remaining two positions for the cyan cups. Hence, the cups may be distributed without restriction in
$$\binom{6}{2, 2, 2} = \binom{6}{2}\binom{4}{2}\binom{2}{2} = \binom{6}{2}\binom{4}{2}$$
distinguishable ways.
From these distributions, we must exclude those in which one or more cups is placed on a saucer of the same color.
Let $A_i$ be the event that an azure cup is placed on the $i$th azure saucer; let $B_i$ be the event that a blue cup is placed on the $i$th blue saucer; let $C_i$ be the event that a cyan cup is placed on the $i$th cyan saucer.
$|A_1|$: Since an azure cup is placed on the first azure saucer, there are two blue cups, two cyan cups, and one azure cup left to distribute to the remaining five saucers. They can be distributed in
$$\binom{5}{1, 2, 2} = \binom{5}{1}\binom{4}{2}\binom{2}{2} = \binom{5}{1}\binom{4}{2}$$
distinguishable ways. By symmetry,
$$|A_1| = |A_2| = |B_1| = |B_2| = |C_1| = |C_2|$$
$|A_1 \cap A_2|$: Since both azure cups have been placed on azure saucers, there are two blue and two cups left to distribute to the remaining four saucers. They can be distributed in
$$\binom{4}{2, 2} = \binom{4}{2}\binom{2}{2} = \binom{4}{2}$$
ways. By symmetry,
$$|A_1 \cap A_2| = |B_1 \cap B_2| = |C_1 \cap C_2|$$
$|A_1 \cap B_1|$: Since an azure cup has been placed on the first azure saucer and a blue cup has been placed on the first blue saucer, there are two cyan cups, one azure cup, and one blue cup left to distribute to the remaining four saucers. They can be distributed in
$$\binom{4}{1, 1, 2} = \binom{4}{1}\binom{3}{1}\binom{2}{2} = \binom{4}{1}\binom{3}{1}$$
distinguishable ways. By symmetry,
$$|A_1 \cap B_1| = |A_1 \cap B_2| = |A_1 \cap C_1| = |A_1 \cap C_2| = |A_2 \cap B_1| = |A_2 \cap B_2| = |A_2 \cap C_1| = |A_2 \cap C_2| = |B_1 \cap C_1| = |B_1 \cap C_2| = |B_2 \cap C_1| = |B_2 \cap C_2|$$
$|A_1 \cap A_2 \cap B_1|$: Since both azure cups have been placed on azure saucers and a blue cup has been placed on the first blue saucer, there are two cyan and one blue cup left to distribute to the remaining three saucers. They can be distributed in
$$\binom{3}{1, 2} = \binom{3}{1}\binom{2}{2} = \binom{3}{1}$$
distinguishable ways. By symmetry,
$$|A_1 \cap A_2 \cap B_1| = |A_1 \cap A_2 \cap B_2| = |A_1 \cap A_2 \cap C_1| = |A_1 \cap A_2 \cap C_2| = |A_1 \cap B_1 \cap B_2| = |A_2 \cap B_1 \cap B_2| = |A_1 \cap C_1 \cap C_2| = |A_2 \cap C_1 \cap C_2| = |B_1 \cap B_2 \cap C_1| = |B_1 \cap B_2 \cap C_2| = |B_1 \cap C_1 \cap C_2| = |B_2 \cap C_1 \cap C_2|$$
$|A_1 \cap B_1 \cap C_1|$: Since an azure cup has been placed on the first azure saucer, a blue cup has been placed on the first blue saucer, and a cyan cup has been placed on the first cyan saucer, we have an azure cup, a blue cup, and a cyan cup to distribute to the remaining three saucers. They can be distributed in
$$\binom{3}{1, 1, 1} = \binom{3}{1}\binom{2}{1}\binom{1}{1} = 3!$$
distinguishable ways. By symmetry,
$$|A_1 \cap B_1 \cap C_1| = |A_1 \cap B_1 \cap C_2| = |A_1 \cap B_2 \cap C_1| = |A_1 \cap B_2 \cap C_2| = |A_2 \cap B_1 \cap C_1| = |A_2 \cap B_1 \cap C_2| = |A_2 \cap B_2 \cap C_1| = |A_2 \cap B_2 \cap C_2|$$
$|A_1 \cap A_2 \cap B_1 \cap B_2|$: Since both azure cups have been placed on azure saucers and both blue cups have been placed on blue saucers, there are two cyan cups to distribute to the remaining two saucers. They can be distributed in
$$\binom{2}{2}$$
distinguishable ways. By symmetry,
$$|A_1 \cap A_2 \cap B_1 \cap B_2| = |A_1 \cap A_2 \cap C_1 \cap C_2| = |B_1 \cap B_2 \cap C_1 \cap C_2|$$
$|A_1 \cap A_2 \cap B_1 \cap C_1|$: Since both azure cups have been placed on azure saucers, a blue cup has been placed on the first blue saucer, and a cyan cup has been placed on the first cyan saucer, we have one blue cup and one cyan cup to distribute to the remaining two saucers. They can be distributed in
$$\binom{2}{1, 1} = \binom{2}{1}\binom{1}{1} = 2!$$
distinguishable ways. By symmetry,
$$|A_1 \cap A_2 \cap B_1 \cap C_1| = |A_1 \cap A_2 \cap B_1 \cap C_2| = |A_1 \cap A_2 \cap B_2 \cap C_1| = |A_1 \cap A_2 \cap B_2 \cap C_2| = |A_1 \cap B_1 \cap B_2 \cap C_1| = |A_1 \cap B_1 \cap B_2 \cap C_2| = |A_2 \cap B_1 \cap B_2 \cap C_1| = |A_2 \cap B_1 \cap B_2 \cap C_2| = |A_1 \cap B_1 \cap C_1 \cap C_2| = |A_1 \cap B_2 \cap C_1 \cap C_2| = |A_2 \cap B_1 \cap C_1 \cap C_2| = |A_2 \cap B_2 \cap C_1 \cap C_2|$$
$|A_1 \cap A_2 \cap B_1 \cap B_2 \cap C_1|$: Since both azure cups have been placed on azure saucers, both blue cups have been placed on blue saucers, and a cyan cup has been placed on the first azure saucer, the remaining cyan cup must be placed on the remaining saucer. There is one way to do this. By symmetry,
$$|A_1 \cap A_2 \cap B_1 \cap B_2 \cap C_1| = |A_1 \cap A_2 \cap B_1 \cap B_2 \cap C_2| = |A_1 \cap A_2 \cap B_1 \cap C_1 \cap C_2| = |A_1 \cap A_2 \cap B_2 \cap C_1 \cap C_2| = |A_1 \cap B_1 \cap B_2 \cap C_1 \cap C_2| = |A_2 \cap B_1 \cap B_2 \cap C_1 \cap C_2|$$
$|A_1 \cap A_2 \cap B_1 \cup B_2 \cap C_1 \cap C_2|$: Since both azure cups have been placed on azure saucers, both blue cups have been placed on blue saucers, and both cyan cups have been placed on cyan saucers, there are no cups left to distribute. There is one way to do this.
Thus, by the Inclusion-Exclusion Principle, the number of ways to distribute the cups so that no cup is on a saucer of the same color is
$$\binom{6}{2}\binom{4}{2} - 6\binom{5}{1}\binom{4}{2} + 3\binom{4}{2} + 12\binom{4}{1}\binom{3}{1} - 12\binom{3}{1} - 8 \cdot 3! + 3 \cdot 1 + 12 \cdot 2! - 6 \cdot 1 + 1$$
Hence, the probability no cup is on a saucer of the same color is
$$\frac{\binom{6}{2}\binom{4}{2} - 6\binom{5}{1}\binom{4}{2} + 3\binom{4}{2} + 12\binom{4}{1}\binom{3}{1} - 12\binom{3}{1} - 8 \cdot 3! + 3 \cdot 1 + 12 \cdot 2! - 6 \cdot 1 + 1}{\binom{6}{2}\binom{4}{2}}$$
Let $n_k$ be the number of $k$-cycles in the disjoint cycle decomposition of $\sigma$. Then
$k \in \{1,2,4,7,14,28\}$
$n_{28}=0$ because $28>14$
$n_{14}=0$ because a $14$-cycle is not even
$n_4 \ge 1$
$n_7 \ge 1$
$n_1 + 2n_2 + 4n_4 +7n_7 = 14$
The last equation has no solutions if $n_1=0$. Thus $n_1\ge1$, as required.
Actually, the only solutions are
$(n_1,n_2,n_4,n_7)=(1,1,1,1)$ and $(3,0,1,1)$.
However, since $\sigma$ is even, we must have $n_2+n_4$ even and so the only solution is $(n_1,n_2,n_4,n_7)=(1,1,1,1)$.
Best Answer
I think that A.4 is not necessary to compute $c_1,...,c_6$.
If $$ \mathcal{A}=\{0.15,0.15,0.05,0.05,0.05,0.02\}\\ \mathcal{B}=\{0.15,0.15,0.11,0.05,0.05,0.02\}$$ then $$c_1=0.3,c_2=0.2,c_3=0.17,c_4=0.16,c_5=0.1,c_6=0.07$$ where $$\begin{cases}a_1=0.15,a_2=0.05,a_3=0.15,a_4=0.05,a_5=0.05,a_6=0.02, \\\\b_1=0.15,b_2=0.15,b_3=0.02,b_4=0.11,b_5=0.05,b_6=0.05\end{cases}$$ or $$\begin{cases}a_1=0.15,a_2=0.15,a_3=0.02,a_4=0.05,a_5=0.05,a_6=0.05, \\\\b_1=0.15,b_2=0.05,b_3=0.15,b_4=0.11,b_5=0.05,b_6=0.02\end{cases}$$
Proof :
If we fix the order of the elements of $\mathcal{B}$ as $(0.15,0.15,0.11,0.02,0.05,0.05)$, then the only possibilities for the order of the elements of $\mathcal{A}$ to consider are
$(0.15,0.05,0.05,0.02,0.15,0.05)$ for which $0.2$ appears more than once as a sum.
$(0.15,0.05,0.02,0.05,0.15,0.05)$ for which $0.2$ appears more than once as a sum.
$(0.15,0.05,0.05,0.05,0.15,0.02)$ for which $0.2$ appears more than once as a sum.
$(0.15,0.02,0.05,0.05,0.15,0.05)$ works.
$(0.15,0.05,0.05,0.15,0.05,0.02)$ works.
$(0.15,0.05,0.15,0.05,0.05,0.02)$ for which $0.07$ appears more than once as a sum.
$(0.05,0.02,0.05,0.15,0.15,0.05)$ for which $0.2$ appears more than once as a sum.
$(0.05,0.02,0.15,0.05,0.15,0.05)$ for which $0.2$ appears more than once as a sum.