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.
 A: There's a term for this type of problem/rearrangement - it is known as a derangement. The derangement of $n$ objects is typically denoted $!n$ (not to be confused with $n!$ - the former is the "subfactorial"), and it can be shown that
$$!n=n!\sum _{i=0}^{n}{\frac {(-1)^{i}}{i!}}$$
(In this case, $n=4$ - four cups/saucers.) 
This will give you the number of valid arrangements meeting your criteria. Then you can just divide that by the total number of arrangements, which you probably can gather to be $4!$.
Thus, you seek
$$\frac{!4}{4!}$$
which gives you the probability. Minor note - notice, you can factor out the $n!$ in summation of the numerator for an even simpler expression, 
$$\frac{!n}{n!} = \sum_{k=0}^n \frac{(-1)^k}{k!}$$
A: 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*}
