<s>[INST] What is the probability that no cup is on a saucer of the same colour? A tea set comprises four cups and saucers in four distinct colours. 
If the cups are placed at random on the saucers, what is the probability that no cup is on a saucer of the same colour?
MY ATTEMPT
As far as I understand, the given event's probability is given by
\begin{align*}
\textbf{P}(C^{c}_{1}\cap C^{c}_{2}\cap C^{c}_{3}\cap C^{c}_{4}) = 1 - \textbf{P}(C_{1}\cup C_{2}\cup C_{3}\cup C_{4})
\end{align*}
where the last expression can be calculated using the inclusion-exclusion principle based on the knowledge of the probabilities
\begin{cases}
\textbf{P}(C_{i})\quad\text{for}\quad1\leq i \leq 4\\
\textbf{P}(C_{i}\cap C_{j})\quad\text{for}\quad 1\leq i < j \leq 4\\
\textbf{P}(C_{i}\cap C_{j}\cap C_{k})\quad\text{for}\quad 1 \leq i < j < k \leq 4\\
\textbf{P}(C_{1}\cap C_{2}\cap C_{3}\cap C_{4})\\
\end{cases}
which can be easily obtained.
 [/INST] There are $4!$ ways to arrange the cups on the saucers.  From these, we wish to exclude those arrangements in which a cup is placed on a saucer of its own color.
If cup $C_i$ sits on saucer $S_i$, there are $3!$ ways to arrange the remaining three cups on the remaining three saucers, so $|C_i| = 3!$, $1 \leq i \leq 4$.
If cups $C_i$ sits on saucer $S_i$ and cup $C_j$ sits on saucer $S_j$, then there are $2!$ ways to arrange the remaining two cups on the remaining two saucers, so $|C_i \cap C_j| = 2!$, $1 \leq i < j \leq 4$.
If cup $C_i$ sits on saucer $S_i$, cup $C_j$ sits on saucer $S_j$, and cup $C_k$ sits on saucer $S_k$, then there is one way to place the remaining cup on the remaining saucer, so $|C_i \cap C_j \cap C_k| = 1!$, $1 \leq i < j < k \leq 4$.
There is only one way to place each cup on its own saucer, so $|C_1 \cap C_2 \cap C_3 \cap C_4| = 0!$.    
Hence, by the Inclusion-Exclusion Principle, 
\begin{align*}
\Pr&(C_1^C \cap C_2^C \cap C_3^C \cap C_4^C) = 1 - \Pr(C_1 \cup C_2 \cup C_3 \cup C_4)\\
& = 1 - \Pr(C_1) - \Pr(C_2) - \Pr(C_3) - \Pr(C_4) + \Pr(C_1 \cap C_2) + \Pr(C_1 \cap C_3) + \Pr(C_1 \cap C_4) + \Pr(C_2 \cap C_3) + \Pr(C_2 \cap C_4) + \Pr(C_3 \cap C_4) - \Pr(C_1 \cap C_2 \cap C_3) - \Pr(C_1 \cap C_2 \cap C_4) - \Pr(C_1 \cap C_3 \cap C_4) - \Pr(C_2 \cap C_3 \cap C_4) + \Pr(C_1 \cap C_2 \cap C_3 \cap C_4)\\
& = 1 - \binom{4}{1}\Pr(C_1) + \binom{4}{2}\Pr(C_1 \cap C_2) - \binom{4}{3}\Pr(C_1 \cap C_2 \cap C_3) + \binom{4}{4}\Pr(C_1 \cap C_2 \cap C_3 \cap C_4)\\
& = 1 - 4 \cdot \frac{3!}{4!} + 6 \cdot \frac{2!}{4!} - 4 \cdot \frac{1!}{4!} + 1  \cdot \frac{0!}{4!}\\
& = 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24}\\
& = \frac{24 - 24 + 12 - 4 + 1}{24}\\
& = \frac{9}{24}\\
& = \frac{3}{8}
\end{align*}
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