Question

The number of $4$ digit numbers divisible by $3$ that can be formed by $4$ different even digits $\{0,2,4,5,6,8\}$ is ?

• A. $36$
• B. $60$
• C. $30$
• D. $24$

Answer 1

The correct option is A $36$

Since the number is divisible by $3$, the sum of the digits should be multiple of $3$.

$\therefore$The digits can be $0,2,4,6$ or $0,4,6,8$

Hence the required no. of numbers $=(4!-3!)+(4!-3!)=36$

($0$ cannot come in 1st position.)