Question

The sum of the digits in the unit place of all numbers formed with the help of $3,4,5,6$ taken all at a time is

• A. $18$
• B. $432$
• C. $108$
• D. $144$

Answer 1

The correct option is C

$108$

Explanation for the correct answer:

Let us form a four digit number with $3$ in the units place.

The number is of the arrangement $\_\_\_3$ where the blanks can be filled by $4,5,6$.

The blanks can be filled in $3!$ ways.

Hence the sum of the of digits in units place is $3\times 3!$

Now, we form a four digit number with $4$ in the units place.

The number is of the arrangement $\_\_\_4$ where the blanks can be filled by $3,5,6$.

The blanks can be filled in $3!$ ways.

Hence the sum of the of digits in units place is $4\times 3!$

Similarly, the sum of the digits of the numbers with $5$ in units place is $5\times 3!$.

Similarly, the sum of the digits of the numbers with $6$ in units place is $6\times 3!$.

Hence, the sum of all digits in the unit place $=\left(3\times 3!\right)+\left(4\times 3!\right)+\left(5\times 3!\right)+\left(6\times 3!\right)$

$=6(3+4+5+6)$

$=108$

Hence, the sum of all digits in the units place formed by using the numbers $3,4,5,6$ is $108$.

Hence, option C is the correct answer.