Question

The sum of all $4$ digit numbers that can be formed by using the digits $2,4,6,8$ (repetition of digits not allowed) is

• A. $133320$
• B. $123330$
• C. $113230$
• D. $323430$

Answer 1

Answer: (A)

There are $4!=24$ numbers. Each digit occurring $3!=6$ times, in the unit's, ten's, hundred’s and thousand's places. We note that $6(2+4+6+8)=120$. Thus in the over all sum there will be $120$ units, $120$ tens, $120$ hundreds and $120$ thousands.
$\therefore$ The required sum $=120(1+10+10^2+10^3)$
$=120\times 1111=133320$.