Question

How many three-digit numbers that are divisible by 5 can be formed using the digits 0, 2, 3, 5, 7, if no digit occurs more than once in each number

• A. 21
• B. 20
• C. 18
  
• D. 15

Answer 1

The correct option is A 21

Since each desired number is divisible by $5$, so we must have $0or5$ at the unit place.
Case 1, when $0$ is at the units place
Then the hundreds and tens place can be filled up by remaining $4$ digits in $\frac{4!}{(4-2)!}=\frac{4!}{2!}=4\times 3=12$ ways
Case 2, when $5$ is at the units place
Then the ten's digit can be filled by the digits $0,2,3,7$ in $4$ ways
And the ${100}^{\prime }s$ digit can be filled by any of the remaining $2$ digits in $2$ ways as $0$ cannot take Hundred's place, then number wont be a $3$-digit number then.
So, total number ways $=12+4+2=18$ ways