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

• A. $44$
• B. $53$
• C. $67$
• D. $265$

Answer 1

The correct option is B $53$

There are $2$ possibilities
$\left(1\right)$ If the card number $2$ goes in the envelope $1$,
then it is derangement case of $4$ things which can be done in $4!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!})=9$
$\left(2\right)$ If card number $2$ doesn’t go in the envelope $1$,
then it is derangement of case $5$ things which can be done in $5!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!})=44$

Hence, total ways $=9+44=53$