Question

The sum of the digits at the tens place of all four digit numbers that can be formed with the help of $3,4,5,6$ taken all at a time is (repetition not allowed):

• A. $432$
• B. $108$
• C. $36$
• D. $18$

Answer 1

The correct option is B $108$

Let us assume that "3" is at tens place

_ _ $\underset{–}{3}$ _

No. of numbers having 3 at tens place

$=3!=6$

similarly,

No. of numbers having $4,5,6$ at tens place are $6,6,6$ respectively

Now sum of numbers at tens place

$=3\left(6\right)+4\left(6\right)+5\left(6\right)+6\left(6\right)$

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

$=6\left(18\right)$

$=108$