Question

The sum of all the numbers of four different digit 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

The correct option is C $38664$

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
$=(6\times 0)+(4\times 1)+(4\times 2)+(4\times 3)=24.$

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
$=(6\times 1)+(6\times 2)+(6\times 3)=36$

Hence, the required sum is $(36\times 1000)+(24\times 100)+(24\times 10)+24=38664$

$\text{Alternate Solution}$
$\text{sum of numbers}=\frac{n}{2}(a+l)$
$n=$ number of total possible number
$a=$ smallest number made by digits$(0,1,2,3)$
$l=$ largest number made by digits$(0,1,2,3)$

But if we observe clearly smallest number will be $0123$ that is actually a three digit number so we need to subtract sum of three digit numbers
Total four digit numbers while considering $0$ at first place $=4!=24$
Total three digit numbers $($by$1,2,3)=3!=6$
Now, required sum $=\frac{24}{2}(0123+3210)-\frac{6}{2}(123+321)$
$=3996-1332$
$=38664$