Question

The sum of all the numbers of four different digits that can be made by using the digits $0,1,2$, and $3$ is

• A. $26664$
• B. $39996$
• C. $38664$
• D. none of these

Answer 1

Answer: (C)

The number of numbers with $0$ in the unit’s place is $3!=6$.
The number of numbers with $1$ or $2$ or $3$ in the unit’s place is $3!-2!=4$. Therefore, the sum of the digits in the unit’s place is $6\times 0+4\times 1+4\times 2+4\times 3=24$.
Similarly, for the ten’s and hundred’s places, the number of numbers with $1$ or $2$ in the thousand’s place is $3!$ Therefore, the sum of the digits in the thousand’s place is $6\times 1+6\times 2+6\times 3=36$.
Hence, the required sum is $36\times 1000+24\times 100+24\times 10+24$.