Question

The number of ways in which all the letters of the word $GARDEN$ can be arranged such that no letter is in its original position is

Answer 1

Given word is $GARDEN$
Number of letters is $6$
So, we need to calculate $D_6$
as we know that $D_n=n!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+.....(-1{)}^n\frac{1}{n!})$
Hence, number of derangements $D_6=6!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!})$
$D_6=720\times (1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}-\frac{1}{120}+\frac{1}{720})$
$D_6=360-120+30-6+1$
$D_6=265$