
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
162.9k+ views
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
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
Recently Updated Pages
Fluid Pressure - Important Concepts and Tips for JEE

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

Impulse Momentum Theorem Important Concepts and Tips for JEE

Graphical Methods of Vector Addition - Important Concepts for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

NEET 2025 – Every New Update You Need to Know
