Question

Explain derangement. Write the formula and give examples.

Answer 1

Hint: In this question, we will first understand the meaning of derangement and then we will use examples to understand it better. We will also state the formula for the number of derangement of a set with n objects.

Complete step-by-step answer:
Let us first understand the meaning of derangement in a set.
Derangement can be defined as a permutational arrangement with no fixed point. We can also say that derangement is the permutation of the elements of a certain set in a way that no element of the set appears in their original position.
For example, for the set {1,2,3} the derangements are {2,3,1} and {3,1,2} because in these two arrangements none of the values occupy their original position.
Let us take one more example. Consider a set {1,2,3,4}. The derangement of this set can be {2,1,4,3}, {2,3,4,1}, {2,4,1,3}, {3,1,4,2}, {3,4,1,2}, {3,4,2,1}, {4,1,2,3}, {4,3,1,2} and {4,3,2,1}. We can see that, in any of these sets, 1 is not in the first position, 2 is not in the second position, 3 is not in the third position and 4 is not in the fourth position.
A derangement for a set of n elements is denoted by !n or ${{D}_{n}}$.
The number of derangement of a set with n objects is given by the formula ${{D}_{n}}=n!\sum\limits_{k=0}^{n}{\dfrac{{{\left( -1 \right)}^{k}}}{k!}}$.

Note: Students should note that in a derangement, not even a single element could be in its own position. Students can also check this formula by using it for the set {1,2,3}. Number of elements are 3 so according to the formula,
\[\begin{align}
  & {{D}_{3}}=3!\sum\limits_{k=0}^{3}{\dfrac{{{\left( -1 \right)}^{k}}}{k!}} \\
 & \Rightarrow 3!\left( \dfrac{{{\left( -1 \right)}^{0}}}{0!}+\dfrac{{{\left( -1 \right)}^{1}}}{1!}+\dfrac{{{\left( -1 \right)}^{2}}}{2!}+\dfrac{{{\left( -1 \right)}^{3}}}{3!} \right) \\
 & \Rightarrow 6\left( 1-1+\dfrac{1}{2}-\dfrac{1}{6} \right) \\
 & \Rightarrow 6\left( \dfrac{2}{6} \right) \\
 & \Rightarrow 2 \\
\end{align}\]
We also found the number of derangements as 2.