Question

The number of ways $10$ letters can be placed in $10$ marked envelopes, so that no letter is in the right envelope are

• A. $10!(\frac{1}{2!}-\frac{1}{3!}+\cdots +\frac{1}{10!})$
• B. $10!(-\frac{1}{2!}+\frac{1}{3!}-\cdots -\frac{1}{10!})$
• C. $(\frac{1}{2!}-\frac{1}{3!}-\cdots +\frac{1}{10!})$
• D. $9!(-\frac{1}{2!}+\frac{1}{3!}-\cdots -\frac{1}{10!})$

Answer 1

The correct option is A $10!(\frac{1}{2!}-\frac{1}{3!}+\cdots +\frac{1}{10!})$

The required number of ways is equal to the number of dearrangements of $10$ objects
Hence, required number of ways
$=10!(\frac{1}{2!}-\frac{1}{3!}+\cdots +\frac{1}{10!})$