Number of $5$ digit numbers which are divisible by $5$ and each number containing the digit $5$, digits being all different is equal to $K\left(4!\right)$, then the value of $K$ is

The correct option is B

$168$

Finding the value of $k$:

The number which is divisible by $5$ when the digit at unit place of that number must be either $5$ or $0$

Given that,

Each $5$ digit numbers which are divisible by $5$ must contain the digit $5$

Case $1$: Consider $0$ at unit place

Number of $5$ digit numbers containing digit $5$ in this case $=4\times 8\times 7\times 6=1344$

Case $2$: Consider $5$ at unit place

First digit cannot be the $0$.

Number of $5$ digit numbers in this case $=8\times 8\times 7\times 6=2688$

Therefore, total number of $5$ digit numbers which are divisible by $5$ must contain the digit $5$$=1344+2688=4032$

Now,

$$\begin{gathered} K\left(4!\right)=4032 \\ K\left(4!\right)=168\times 24 \\ K\left(4!\right)=168\left(4!\right) \end{gathered}$$

Equating both side

$K=168$

Hence, the value of $K=168$