Question

The sum of all the numbers of four different digits that can be made by using the digits \[0,{\rm{ }}1,{\rm{ }}2\] and \[3\] is
A. \[64322\]
B. \[48522\]
C. \[38664\]
D. \[1000\]

Answer 1

Hint: In the given question, we need to find the sum of all the numbers of four different digits that can be made by using the digits \[0,{\rm{ }}1,{\rm{ }}2\] and \[3\]. For this, we will find the sum of the digits in the units place, tens place, hundreds place and thousand's place. After that we will get the required sum

Formula used: The following formula used for solving the given question.
The factorial of a number is given by, \[n! = n \times (n - 1) \times .... \times 1\]
Here, \[n\] is a given number.

Complete step by step solution: We know that \[n! = n \times (n - 1) \times .... \times 1\]
Here, the number of numbers with zero in the unit's place is \[3! = 6.\]
Also, the number of numbers with one or two or three in the unit's place is \[3! - 2! = 4.\;\]
Thus, we can say that the sum of the digits in the unit's place is \[6 \times 0 + 4 \times 1 + 4 \times 2 + 4 \times 3 = 24.\]
Similarly, for the tens and hundreds place, the number of numbers with one or two in the thousand's place is \[3!\].
Hence, the sum of the digits in the thousand's place is \[6 \times 1 + 6 \times 2 + 6 \times 3 = 36\].
Thus, the required sum is \[36 \times 1000 + 24 \times 100 + 24 \times 10 + 24 = 38664\]

Thus, Option (C) is correct.

Additional Information: A positive integer's factorial is the result of multiplying all the negative integers by the positive integer.

Note: Many students make mistakes in the calculation part as well as finding the sum of the digits in respective places of a number. This is the only way through which we can solve the example in the simplest way. Also, it is essential to find correct result of the required su