Derangement : arrangement where no element is in its designated position
Number of derangement can be calculated easily using the principle of mutual exclusion and inclusion (http://math.mit.edu/~fox/MAT307-lecture04.pdf )
In this question, we take the designated position of ball $B_i$ as $C_i$.
We need to find all possible arrangements where no $B_i$ is in its designated cell $C_i$.
Lets take $5$ arrangements for positions $1,2,3,4$ and $5$ where, in each of them, one element is in its designated position and denote them as $S_i$.
$$\begin{array}{|c|c|c|c|} \hline \text {$1$} & \text{ $x$}& \text{$x$} & \text{$x$} & \text{$x$} \\\hline \end{array} \rightarrow{S_1}$$
$|{S_1}| = 4!$ as the other 4 balls can be arranged in $4!$ ways inside $S_1$
Similarly, we have $S_2$ which is the arrangement of all balls with $B_2$ in its designated position.
$$\begin{array}{|c|c|c|c|} \hline \text {$x$} & \text{ $2$}& \text{$x$} & \text{$x$} & \text{$x$} \\\hline \end{array} \rightarrow{S_2}$$
We have $|{S_2}| = 4!$ similar to $|S_1|$ and likewise $|{S_i}| = 4!$ .
We have total number of non-derangement $=\ |{S_1}\cup{S_2}\cup{S_3}\cup{S_4}\cup{S_5}| $
According to the principle of Mutual Inclusion and Exclusion,
$|{S_1}\cup{S_2}\cup{S_3}\cup{S_4}\cup{S_5}| = \Sigma|{S_i}|-\Sigma|{S_i\cap{S_j}}|+\Sigma|S_i\cap{S_j}\cap{S_k}|-\ldots \longrightarrow (A)$
$|S_1\cap{S_2}| = $number of arrangements where $B_1$ and $B_2$ are in their designated positions.
$$\begin{array}{|c|c|c|c|} \hline \text {$1$} & \text{ $2$}& \text{$x$} & \text{$x$} & \text{$x$} \\\hline \end{array}$$
This can be done in $3!$ ways.
Similarly all the other two intersections,$ |S_i\cap {S_j}| = 3!$ and three intersections,$|S_i\cap{S_j}\cap{S_k}| = 2! $ and similarly all.
Substituting in equation $(A),$
$|{S_1}\cup{S_2}\cup{S_3}\cup{S_4}\cup{S_5}|= 5\times 4!-\binom{5}{2}\times 3!+\binom{5}{3}\times 2!-\binom{5}{4}\times 1!+\binom{5}{5}\times 0!=76$ (equals number of non derangements)
Derangements = all arrangements - non derangements
$= 5! - 76 = 44$
Correct Answer: A
