Question

The number of ways in which all the letters of the word $HOUSE$ can be arranged such that no letter will occur in its original position is $11k$, then the value of $k$ is

Answer 1

Given word is $HOUSE$
Number of letters is $5$
So, we need to calculate dearangement of $5$ letters $D_5$
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 $D_5=5!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!})$
$D_5=120\times (1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}-\frac{1}{120})$
$D_5=60-20+5-1$
$D_5=44=11k$
$k=4$