Question

A person writes letters to six friends and addressed the corresponding envelopes. Let "k" be the number of ways the letters be placed in the envelopes so that at least two of them are in the wrong envelopes. Find sum of digits of "k" ?

Answer 1

$\left(i\right)$ The number of ways in which at least two of them are in the wrong envelopes
$=\sum _{r=2}^6{}^n{C}_{n-r}{D}_r$
$={}^n{C}_{n-2}{D}_2+{}^n{C}_{n-3}{D}_3+{}^n{C}_{n-4}{D}_4+{}^n{C}_{n-5}{D}_5+{}^n{C}_{n-6}{D}_6$
Here $n=6$
$={}^6{C}_4.2!(1-\frac{1}{1!}+\frac{1}{2!})+{}^6{C}_3.3!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!})$
$+{}^6{C}_2.4!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!})$
$+{}^6{C}_1.5!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!})$
$+{}^6{C}_0.6!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!})$
$=15+40+135+264+265$
$=719$