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)

$133320$

Answer 2

There are $4! = 24$ numbers. Each digit occurring $3! = 6$ times, in the unit's, ten's, hundred?? 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 $= 120 (1 + 10 + 10^2 + 10^3)$ $= 120 \times 1111 = 133320$.