Question

A person writes letters to six friends and addresses the corresponding envelopes. Let $x$ be the number of ways so that at least two of the letters are in wrong envelopes and $y$ be the number of ways so that all the letters are in wrong envelopes, then $x-y=$

• A. 716
• B. 454
• C. 265
• D. 0

Answer 1

The correct option is B 454

$x$: Number of ways of getting 2 wrong OR 3 wrong OR 4 wrong OR 5 wrong OR 6 wrong
$y$: Number of ways of getting 6 wrong
$x-y$: Number of ways of getting 2 wrong OR 3 wrong OR 4 wrong OR 5 wrong
In general, number of ways of getting r wrong: ${}^n{C}_r\times r!\times (\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}...{(-1)}^r\frac{1}{r!})$
Thus, $X-Y{=}_2^{6}C\times 2!\times \left(\frac{1}{2!}\right){+}_3^{6}C\times 3!\times (\frac{1}{2!}-\frac{1}{3!}){+}_4^{6}C\times 4!\times (\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}){+}_5^{6}C\times 5!\times (\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!})=15+40+135+264=454$
Hence, (b) is correct.