Question

The sum of all $4-$digit even numbers that can be formed from the digits $1,2,3,4,5$ (without repetition) is:

• A. $1,58,994$
• B. $1,59,984$
• C. $1,59,894$
• D. $1,59,884$

Answer 1

The correct option is B $1,59,984$

Given digits are $1,2,3,4,5$

Let the $4-$digit number be $ABCD$

$ABCD=1000A+100B+10C+D$

$D$ can be $2$ or $4$ only.

Therefore, total number of $4-$digit even numbers are

$4\times 3\times 2\times 2=48$

$2$ appears at unit's place exactly $24$ times, and $4$ appears at unit's place exactly $24$ times

Remaining places can be filled with $12$ times appearance of each of $1,3,5$ and $6$ times appearance of each of $2,4$.

Therefore,

The sum of unit's place of all numbers is

$24\times 2+24\times 4=144$

The sum of Ten's place of all numbers is

$6\times 2+6\times 4+12\times 5+12\times 3+12\times 1=144$

The sum of Hundred's place of all numbers is

$6\times 2+6\times 4+12\times 5+12\times 3+12\times 1=144$

The sum of Thousand's place of all numbers is

$6\times 2+6\times 4+12\times 5+12\times 3+12\times 1=144$

Now, the sum of all $4-$digit even numbers

Sum $=1000\sum A+100\sum B+10\sum C+\sum D$

$\Rightarrow$ Sum $=1000\times 144+100\times 144+10\times 144+144$

$\Rightarrow$ Sum $=(1000+100+10+1)\times 144$

$\Rightarrow$ Sum $=159984$

Hence, option B.