Problem 33 - Algebra for JEE ADVANCED

Question

The sum of all the numbers of four different digits that can be made by using the digits \(0,1,2\) and 3 is a. 26664 b. 39996 c. 38664 d. none of these

Step-by-Step Solution

Answer

The sum of all possible numbers is 119988, which doesn't match any options. Hence, the answer is d. none of these.

Step 1: Understanding the Digits

We're given the digits 0, 1, 2, and 3 and asked to find the sum of all four-digit numbers that can be formed using these digits without repetition. Importantly, since a four-digit number cannot start with 0, we should exclude the numbers starting with 0 from our sum.

Step 2: Calculating Total Numbers with Each Digit

For any fixed digit in the thousands place (1, 2, or 3), the remaining three digits can be arranged in the hundreds, tens, and units places. Therefore, the number of combinations is calculated as: \[ 3 imes 3! = 3 imes 6 = 18 \]Each digit (1, 2, or 3) will appear in the thousands place for 18 numbers.

Step 3: Sum Contributions from Thousands Place

When each of 1, 2, and 3 appears in the thousands place of 18 numbers, their contribution is:\[ 1000 imes (1 + 2 + 3) imes 18 = 1000 imes 6 imes 18 = 108000 \]

Step 4: Sum Contributions from Each Place

Each digit will statistically appear equally in each place value over the entire set of numbers. Since each digit appears an equal number of times (18 times) in each position including hundreds, tens, and units:- For hundreds place contribution:\[ 100 imes (1 + 2 + 3) imes 18 = 100 imes 6 imes 18 = 10800 \]- For tens place contribution:\[ 10 imes (1 + 2 + 3) imes 18 = 10 imes 6 imes 18 = 1080 \]- For units place contribution:\[ 1 imes (1 + 2 + 3) imes 18 = 1 imes 6 imes 18 = 108 \]

Step 5: Calculating Total Sum

Adding all contributions together:\[ 108000 + 10800 + 1080 + 108 = 119988 \]

Step 6: Finding the Sum for Four-digit Numbers Involving Zero

Since zero cannot appear in the thousands place, we exclude it initially. When zero appears in the other positions, the effect is already implicitly considered in arranging 1, 2, and 3 as each position influences the sum by its place value.

Key Concepts

Permutations

When working with problems related to permutations, it's pivotal to understand that permutations are all about arranging things in a specific order. In this scenario, we have four digits 0, 1, 2, and 3. However, when forming four-digit numbers, 0 cannot be at the start, reducing our leading digit choices to 1, 2, or 3. The permutations of the remaining three digits can then be arranged in any order to complete the number. Thus, we calculate permutations using the formula for factorial, denoted as \( n! \). For three digits, the number of ways to arrange them would be \( 3! = 6 \). Combining this with the three choices for the first digit, we have a total of \( 3 \times 3! = 18 \) possible four-digit numbers. Recognizing this component is key in understanding where most permutations problems start and how you can manipulate digits within constraints.

Digit Arrangements

Digit arrangements in this context refer to the order in which the digits 0, 1, 2, and 3 are placed within a number to create all possible unique combinations. Strategically, since a number cannot start with 0, we are primarily interested in arrangements where 1, 2, or 3 lead the number.For each choice of leading digit, the other three digits can take any of the remaining positions, meaning you can have different orderings for each fixed starting point. The formula \( 3! \) helps us calculate the permutations of these remaining digits. Each permutation influences which numbers are valid, and we systematically follow this logic to enumerate all digit arrangements possible under the stated conditions.Understanding these arrangements helps in computational situations where digits in different places yield varied outcomes affecting the total sum.

Place Value Contributions

Place value contributions are a crucial aspect of calculating the sum of numbers formed by digit arrangements. Each digit within a number carries a place value—thousands, hundreds, tens, or units—significantly affecting its overall contribution.For instance, if a digit is placed in the thousands position, its impact is much more significant compared to the tens or units places. In our problem, each of the non-zero digits (1, 2, and 3) appears 18 times in each position due to our permutations. To calculate the contribution to the overall sum, we multiply the place value by the sum of the digits repeated across all arrangements. For the thousands place the contribution is: - \( 1000 \times (1 + 2 + 3) \times 18 \).Similarly, computations are done for hundreds, tens, and units. As the place value decreases, so does the contribution to the total sum. Approaching the problem through place value analysis enables one to calculate large cumulative sums efficiently by understanding how each digit and its position alters the overall total.