Question

Number of $5$ digit numbers which are divisible by $5$ and each containing the digit $5$, digits being all different is equal to $k(4!)$, the value of $k$ is

• A. $84$
• B. $168$
• C. $188$
• D. $208$

Answer 1

Answer: (B)

$168$

Each contain the digit $5$, so we need to choose remaining $4$ digits out of $9$ digits.

Now two cases- $\left(i\right)0$ is not one of the $4$ digits, $\left(ii\right)0$ is one of the $4$ digits.

Case $\left(i\right)$ As $0$ is not included so for number to be divisible by $5$ must be the last digit and hence remaining $4$ digits can be choosen as $8\times 7\times 6\times 5$

Case $\left(ii\right)$ As $0$ is included so two choices for the last digit $0$ or $5$.

First, Considering $0$ as last digit, $5$ can be put on any of the $4$ places, leaving $3$ places and $8$ different digits that can be put in this $3$ places as $8\times 7\times 6\to$ Total=$4\times 8\times 7\times 6$

Second, Considering $5$ as last digit, $0$ can't be put in first position because than it will be $4$ digit number, So for first digit there are $8$ choices, now $0$ can be placed at any $3$ remaining positions, leaving $2$ places and $7$ different digits that can be put in this $2$ places as $7\times 6\to$ Total=$8\times 3\times 7\times 6$

$\therefore$ Total ways=$8\times 7\times 6\times 5+4\times 8\times 7\times 6+8\times 3\times 7\times 6=12\times 8\times 7\times 6=24\times 4\times 7\times 6=4!\times 168$

So $k=168$.

Hence $\left(B\right)$.