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 exctly 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. $264$
• B. $268$
• C. $53$
• D. $67$

Answer 1

Answer: (C)

$53$

Answer 2

Total number of ways to place $6$ cards in wrong envelopes $=6!\left(\frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}\right)$ $=\left(720-720+360-120+30-6+1\right)$ $=265$ If card numbered $(1)$ is placed in $2$, the no. of ways $= 265/5 = 53$