Problem 93 - A Complete Resource Book in Mathematics for JEE Main

Question

The sum of all numbers greater than 1000 formed by using the digits \(0,1,2,3\), no digit being repeated in any number, is (A) 38664 (B) 48664 (C) 58664 (D) None of these

Step-by-Step Solution

Answer

The sum is 47988, so the answer is (D) None of these.

Step 1: Understanding the Task

We need to find the sum of all 4-digit numbers greater than 1000 that can be formed using the digits 0, 1, 2, and 3 without repeating any digit.

Step 2: Identifying Eligible Number Combinations

To form numbers greater than 1000, the first digit must be either 1, 2, or 3 (the first digit cannot be 0). This yields three options for the first digit.

Step 3: Forming Numbers with Each First Digit

For each choice of the first digit, we permute the remaining digits to form 3-digit numbers. Specifically, once a first digit is fixed, there are 3 remaining digits, allowing for permutations in the remaining 3 positions.

Step 4: Counting Permutations for Each Case

For each choice of the first digit, there are 3! = 6 permutations of the remaining three digits. Thus, 3 (choices for the first digit) * 6 (permutations for each choice) = 18 numbers overall.

Step 5: Calculating Contribution of Each Digit Position

Every digit (0, 1, 2, 3) appears in each position (thousands, hundreds, tens, and units) exactly the same number of times. Since each digit appears once in each of the 6 permutations for each first digit choice, this happens a total of 6 times for each of the three possible first digits, leading to 6 * 3 = 18 times each digit appears in each position.

Step 6: Summing Up Using Positional Values

Calculating each digit's contribution: - Thousands position: 1, 2, 3 appear 6 times each (3 choices, 6 permutations in each). Contribution = 1000 * 6 * (1 + 2 + 3) = 36000. - Hundreds, Tens, and Units positions: Each digit appears 18 times (6 times for 3 numbers). Contribution = (18 * (1 + 2 + 3) * 100) + (18 * (1 + 2 + 3) * 10) + (18 * (1 + 2 + 3)). That's 10800 + 1080 + 108 = 11988.

Step 7: Total Sum Calculation

Combining contributions across all positions, the total sum is 36000 + 11988 = 47988.

Step 8: Comparing With Options Given

Upon calculation, none of the provided options match the sum calculated above. Therefore, the answer is (D) None of these.

Key Concepts

Combinatorial Counting

Combinatorial counting is a fundamental concept in mathematics that deals with finding the number of ways to arrange or choose objects. This is achieved by finding permutations and combinations for different scenarios. In our problem, we are dealing with permutations, which involve arranging a set of distinct objects in a specific order.

When using the digits 0, 1, 2, and 3 to form numbers, each digit must be used exactly once unless stated otherwise. This ensures that no repetition happens in the process of number formation. Here, we're specifically counting the permutations where the first digit (thousands place) is non-zero, allowing numbers to be greater than 1000.

Using the principle of permutations, for n different objects, the number of permutations is given by n-factorial, denoted as \(n!\). For any choice of the first digit in our problem, the arrangement of the remaining three digits is given by \(3! = 6\), implying six different numbers for each case. By understanding this concept, we can seamlessly approach the task of forming valid numbers above a certain threshold.

Digit Sum

Digit sum involves calculating the total contribution to the sum by each digit when used in forming numbers. In problems like these, it’s crucial to understand how positional values contribute to the overall number.

Every digit takes turns occupying each position (thousands, hundreds, tens, and units) a set number of times. When digits are permuted, each digit shows up in these positions repeatedly—but evenly distributed across the permutations. In our specific exercise, digits 1, 2, and 3 appear six times in the thousands position across all permutations. This is because 0 cannot be the leading digit in a 4-digit number that’s greater than 1000.

By computing the digit contribution in different positional values—such as thousands, hundreds, tens, and units—the calculated sum for any required group of numbers can be achieved. For instance, the thousands' position contributes significantly more (1000 times the digit value) than the tens or units positions, emphasizing the importance of weighing each digit's placement carefully.

Four-Digit Number Formation

The formation of four-digit numbers using specific digits is dictated by a set of rules. When forming numbers like those in our problem, where digits 0, 1, 2, and 3 are used, some conditions must be observed to ensure numbers meet the criteria (greater than 1000).

The key constraint here is the restriction on the first digit. Since numbers need to be greater than 1000, the leading digit cannot be 0. This leaves digits 1, 2, or 3 as the only viable options. Each selection of a leading digit constrains the remaining options for the subsequent three digits.

Once the leading digit is chosen, the remaining digits are permuted to fill the hundreds, tens, and units positions. This results in a series of permutations that, when summed up, give the total value of numbers that can be formed. Understanding the rules of number formation—such as non-repetition of digits and limitations on leading zeros—is crucial for solving these types of combinatorial counting problems efficiently.