Question

The sum of the digits in the unit place of all the numbers formed with the help of 3, 4, 5, 6 taken all at a time is:
A. 432
B. 108
C. 36
D. 18

Answer 1

Hint: Form four numbers in such a way that their unit place is 3, 4, 5, 6. Then take the sum of digits in their unit place when it is 3, 4, 5, 6. Take the sum of the four cases.

Complete step-by-step answer:
Given to us are four numbers 3, 4, 5, 6. Thus we need to find the sum of the digits in the unit place of all numbers formed with help of 3, 4, 5, 6.
Thus we need to get four numbers formed 3, 4, 5, 6, and we need to take the sum of unit digits of the number, i.e. \[\underline{{}}\underline{{}}\underline{{}}\underline{x}\].
Now the given 4 digits- 3, 4, 5, 6- have to be used in every number.
Now let us form the four digit number. Let us first place 3 in the unit’s place, i.e. \[\underline{{}}\underline{{}}\underline{{}}\underline{3}\].
The other places can be filled by 4, 5, 6 in 3! ways.
Thus total ways of forming the number with 3 in unit’s place \[=3\times 3!\] ways \[.....(1)\]
Similarly, let us place the digit 4 in the unit’s place, i.e. \[\underline{{}}\underline{{}}\underline{{}}\underline{4}\].
Now the other 3 places can be filled by 3, 5, 6 in 3! ways.
\[\therefore \]Total ways of forming the number with 4 in its units place \[=3!\times 4\] ways \[......(2)\]
Now let us place digit 5 in the units place, i.e. \[\underline{{}}\underline{{}}\underline{{}}\underline{5}\].
\[\therefore \]Total ways of forming the number with 5 in its units place \[=3!\times 5\] ways \[......(3)\]
Let us place digit 6 in the units place, i.e. \[\underline{{}}\underline{{}}\underline{{}}\underline{6}\].
\[\therefore \]Total ways of forming the number with 6 in its units place \[=3!\times 6\] ways \[......(4)\]
\[\therefore \]The sum of digits in unit place when 3 is at their unit place \[=3\times 3!\]
Similarly, when unit place is 4 \[=3!\times 4\]
when unit place is 5 \[=3!\times 5\]
when unit place is 6 \[=3!\times 6\]
Now let us take the sum of equation (1), (2), (3) and (4).
\[\therefore \]The sum of the digits in the unit place of all the numbers formed with the help of 3, 4, 5, 6 taken all at a time is
\[\begin{align}
  & =\left( 3!\times 3 \right)+\left( 3!\times 4 \right)+\left( 3!\times 5 \right)+\left( 3!\times 6 \right) \\
 & =3!\left( 3+4+5+6 \right) \\
 & =3\times 2\times 18=108. \\
\end{align}\]
\[\therefore \]The sum of digits in the unit place of all numbers formed with the help of 3, 4, 5, 6 taken all at a time is 108.
Option B is the correct answer.

Note: We have been given 4 digits 3, 4, 5, 6. It is said that we have to form numbers with the help of these digits. Nothing about repetition or using some numbers twice and all is mentioned. Thus to form the required number you need to take all the 4 digits in all the four cases like equation (1), (2), (3), (4), which we have discussed.