Given, six cards and six envelops are numbered $$1,2,3,4,5$$ and 6.

The number of arrangements when all six

cards are placed in wronge envelops

$$\begin{aligned} & =\left| \!{\underline {\, {6} \,}} \right.\left(1-\frac{1}{\left| \!{\underline {\, {1} \,}} \right.}+\frac{1}{\left| \!{\underline {\, {2} \,}} \right.}-\frac{1}{\left| \!{\underline {\, {3} \,}} \right.}+\frac{1}{\left| \!{\underline {\, {4} \,}} \right.}-\frac{1}{\left| \!{\underline {\, {5} \,}} \right.}+\frac{1}{\left| \!{\underline {\, {6} \,}} \right.}\right) \\ & =\left| \!{\underline {\, {6} \,}} \right.\left[1-\frac{1}{1}+\frac{3}{\left| \!{\underline {\, {3} \,}} \right.}-\frac{1}{\left| \!{\underline {\, {3} \,}} \right.}+\frac{5}{\left| \!{\underline {\, {5} \,}} \right.}-\frac{1}{5}+\frac{1}{\left| \!{\underline {\, {6} \,}} \right.}\right] \\ & =\left| \!{\underline {\, {6} \,}} \right.\left[\frac{2}{\left| \!{\underline {\, {3} \,}} \right.}+\frac{4}{\left| \!{\underline {\, {5} \,}} \right.}+\frac{1}{\left| \!{\underline {\, {6} \,}} \right.}\right] \\ & =2.6 .5 .4+4.6+1 \\ & =240+24+1 \\ & =265 \\ \end{aligned}$$

In the above dearrangement, there are 5 ways in which card number 1 is going wrong envelope i.e, other than envelope number 1. So, when card number 1 going in envelop number 2 is $$\frac{265}{5}=53$$ ways.

Hint:

In the total dearrangement of 6 cards there are 5 ways in which card number 1 is going other than envelope number 1.