Question

The sum of the digits in the units place of all number formed with the help of 3,4,5,6 taken all at a time is
\[
  A.{\text{ 18}} \\
  B.{\text{ 432}} \\
  C.{\text{ 108}} \\
  D.{\text{ 144}} \\
 \]

Answer 1

Hint: In order to solve such a type of problem first individually consider all the digits for the units place and then find the number of combinations possible for each digit as units digit. Then sum up for each digit and number of times of its appearance.

Complete step-by-step answer:
We know that number of arrangements of n digits in n different place is $n!$
We have 3,4,5,6 as the digits to form the number taking all at once.
This means the number will always be a 4 digit number.
Now let us first consider that units place has 3.
When 3 is in units place then we have 3 different numbers 4, 5, 6 which can be arranged in tens, hundreds and thousands place in $3!$ different ways.
$\therefore $ The sum of the digits in units place when 3 is at the units place
$
   = 3! \times 3 \\
   = 6 \times 3 = 18 \\
 $
Similarly
Now let us consider that units place has 4.
When 4 is in units place then we have 3 different numbers 3, 5, 6 which can be arranged in tens, hundreds and thousands place in $3!$ different ways.
$\therefore $ The sum of the digits in units place when 4 is at the units place
$
   = 3! \times 4 \\
   = 6 \times 4 = 24 \\
 $
Similarly
Now let us consider that units place has 5.
When 5 is in units place then we have 3 different numbers 3, 4, 6 which can be arranged in tens, hundreds and thousands place in $3!$ different ways.
$\therefore $ The sum of the digits in units place when 5 is at the units place
$
   = 3! \times 5 \\
   = 6 \times 5 = 30 \\
 $
Similarly
Now let us consider that units place has 6.
When units place has 6 we have 3 different numbers 3, 4, 5 which can be arranged in tens, hundreds and thousands place in $3!$ different ways.
$\therefore $ The sum of the digits in units place when 6 is at the units place
$
   = 3! \times 6 \\
   = 6 \times 6 = 36 \\
 $
Now finally the sum of the digits in the units place of all numbers formed with the help of 3, 4, 5, 6 taken all at a time is
$
   = 3! \times 3 + 3! \times 4 + 3! \times 5 + 3! \times 6 \\
   = 6 \times 3 + 6 \times 4 + 6 \times 5 + 6 \times 6 \\
   = 18 + 24 + 30 + 36 \\
   = 108 \\
 $
Hence, the sum of the digits in the units place of all numbers formed with the help of 3,4,5,6 taken all at a time is 108.
So, option C is the correct option.

Note: In order to solve such types of problems always use the concept of permutation and combination. Whenever such a problem has 0 as a digit and a number has to be formed, always keep in mind that 0 cannot be used as the highest place value digit. Students must remember the formula for the number of arrangements that can be made from n digits.