Question

Six cards and six envelopes are numbered $1,2,3,4,5,6$ and cards are to be placed in envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover the card numbered$1$ is always placed in envelope numbered $2.$ Then the number of ways it can be done is

Answer 1

Derangement: A derangement is a permutation of the elements of a set, such that no element appears in its original position.

The number of derangements of $6$

$$\begin{gathered} =6!(1–\frac{1}{1!}+\frac{1}{2!}–\frac{1}{3!}+\frac{1}{4!}–\frac{1}{5!}+\frac{1}{6!}) \\ =6!(1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}-\frac{1}{120}+\frac{1}{720}) \\ =720\left(\frac{360-120+30-6+1}{720}\right) \\ =360-120+30-6+1 \\ =265 \end{gathered}$$

Out of these derangements, there are five ways in which card numbered $1$ is going wrong.

So, when it is going in envelope numbered $2$,

then remaining number of ways are

$\frac{265}{5}=53$ ways.

Hence, the number of ways it can be done are $53$.