Derangement - Practice Questions & MCQ

If there are n things and n places, one correct place corresponds to each object. Then an arrangement in which none of the objects is in its right place is called a derangement.

The number of ways of doing this is denoted by $D(n)$ (the number of ways of deranging ' $n$ ' objects). The formula for this is

$
\mathrm{D}(\mathrm{n})=\mathrm{n}!\left(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\ldots+(-1)^{\mathrm{n}} \frac{1}{\mathrm{n}!}\right)
$
GOLDEN TIP:

Substituting the value of ' $n$ ' as $1,2,3,4,5,6$ we will get,

$
\begin{aligned}
& D(1)=0 \\
& D(2)=1 \\
& D(3)=2 \\
& D(4)=9 \\
& D(5)=44 \\
& D(6)=265
\end{aligned}
$
A quicker way to find out the total number of possible derangements is just to memorize the above values by heart and use them instantly in the questions.

Example: In how many ways can you form a dancing couple from 3 boys and 3 girls so that no boy dances with his respective girlfriend?

Solution: This is a case of derangement of 3 boys and 3 girls.
The value can be interpreted as $D(3)=2$ ways