Question

The number of ways in which four different letters can be put in the correspondingly five addressed envelopes so that no letter is put in the correct envelope, is?
(A) 6
(B) 7
(C) 8
(D) 9

Answer 1

The given problem of disarrangement can be solved by using the derangement formula which is
Number of derangements $=\left| \\!{\underline {\, n \,}} \right. \times \left( 1-\dfrac{1}{\left| \\!{\underline {\, 1 \,}} \right. }+\dfrac{1}{\left| \\!{\underline {\, 2 \,}} \right. }-\dfrac{1}{\left| \\!{\underline {\, 3 \,}} \right. }+...\pm \dfrac{1}{\left| \\!{\underline {\, n \,}} \right. } \right)$
We then put $n=4$ in the above formula and get our final answer.

Complete step-by-step answer:
There are four different letters and five different envelopes. We need to put the letters in the envelopes in such a manner that none of the letters go into their destined envelope. We can solve this problem by counting the numbers of ways in which letters can go into their destined envelopes and finally add up all the ways and subtract it from the total number of possible arrangements.
Now, we have a commonly used concept for these types of problems, which is known as the derangement concept. This concept deals with the number of ways in which a number of objects can be misarranged.
The formula for derangement is
Number of derangements $=\left| \\!{\underline {\, n \,}} \right. \times \left( 1-\dfrac{1}{\left| \\!{\underline {\, 1 \,}} \right. }+\dfrac{1}{\left| \\!{\underline {\, 2 \,}} \right. }-\dfrac{1}{\left| \\!{\underline {\, 3 \,}} \right. }+...\pm \dfrac{1}{\left| \\!{\underline {\, n \,}} \right. } \right)$
Where, $n$ is the numbers of objects required to be misarranged.
In the given problem, we need to put $4$ letters in the $5$ envelopes in such a manner that none of the letters go into their destined envelope. Thus, here, $n=4$ . Putting this value of $n$ in the formula for derangements, we can write,
Number of derangements can be found as

4 \,}} \right. \times \left( 1-\dfrac{1}{\left| \\!{\underline {\, 1 \,}} \right. }+\dfrac{1}{\left| \\!{\underline {\, 2 \,}} \right. }-\dfrac{1}{\left| \\!{\underline {\, 3 \,}} \right. }+\dfrac{1}{\left| \\!{\underline {\, 4 \,}} \right. } \right)$$ $$\Rightarrow \left| \\!{\underline {\, 4 \,}} \right. \times \left( \dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{24} \right)$$ $$\Rightarrow \left| \\!{\underline {\, 4 \,}} \right. \times \left( \dfrac{12-4+1}{24} \right)$$ $$\begin{aligned} & \Rightarrow 24\times \dfrac{9}{24} \\\ & \Rightarrow 9 \\\ \end{aligned}$$ Therefore, we can conclude that the number of ways in which four different letters can be put in five different envelopes so that no letter is put in the correct envelope is $9$ ways. **Note:** For these types of disarrangement problems, we must remember the derangement formula, otherwise it becomes really hard to calculate all the cases and this becomes prone to error. Students also commit mistakes in considering the value of $n$ which is $4$ here. They mistake it for $5$ .