Problem 91 - A Complete Resource Book in Mathematics for JEE Main

Question

The sum of all the numbers that can be formed by writing all the digits \(3,2,3,4\) only once is (A) 39996 (B) 49996 (C) 57776 (D) None of these

Step-by-Step Solution

Answer

The correct answer is (A) 39996.

Step 1: Understanding the Problem

We need to find the sum of all numbers that can be formed using the digits 3, 2, 3, and 4, where each digit is used exactly once in each number. We must consider how many such numbers can be formed and their possible combinations.

Step 2: Counting Possible Numbers

Since the digits 3 is repeated, we calculate the total number of unique permutations using the formula for permutations of multiset: \(\frac{n!}{p_1!p_2!...p_k!}\). Here, \(n = 4\) and we have 2 repeated digits (3). Thus, the total permutations are \(\frac{4!}{2!} = 12\).

Step 3: Calculating Sum of Each Digit's Contribution by Position

Each digit will appear in each position (thousands, hundreds, tens, units) an equal number of times. Divide the total number of numbers, 12, by the 4 positions: \(12 \div 4 = 3\). Each digit appears 3 times in each position.

Step 4: Calculating Contribution of Each Digit

Calculate the contribution of each digit. For example, digit 3 contributes to 3,000s: \(3 \times 3 \times 1000 = 9000\). Do the calculation for other positions and rest of the digits: \(3 \times 3 \times 100 = 900, 3 \times 2 \times 10 = 60,\) plus contributions from digit 2 and digit 4, summing all individually according to similar contributions.

Step 5: Add All Contributions

Sum the contributions of each digit position: \(3_{thousands} + 4_{thousands} + 3{thousands} + 2_{thousands}\) etc. Combine contributions: (3 as 1000s) + (4 as 1000s) + (2 as 1000s), (3 as 100s) + ... to obtain total sum.

Step 6: Final Calculation

Calculate the cumulative sum of all contributions. Sum digits from positions together as calculated to get the total sum.

Key Concepts

Combinatorics

Combinatorics is a fascinating branch of mathematics that explores the myriad possibilities of arranging a set of items. It helps us understand how to count, arrange, and list items systematically. In the context of permutations, combinatorics asks: 'How many unique ways can we arrange a given set of items or numbers?' When handling sets with repeated elements, as in our exercise involving digits 3, 2, 3, and 4, combinatorics comes into play to calculate arrangements without duplicating effort. The solution involves using formulas designed specifically for such tasks, ensuring all possible permutations are accounted for.

Digits

Handling digits, especially in the context of permutation problems, requires us to appreciate their positional values within numbers. In this exercise, digits 3, 2, 3, and 4 were considered for forming numbers. Each digit must appear exactly once in each number, showing its importance in number formation and permutation. The repeated digit 3 highlights a key area in combinatorics: treating duplicates correctly ensures an accurate count of numbers. Understanding the individual value of each digit at different placements within a number is crucial for calculating sums and permutations accurately.

Number Formation

Number formation is the process of arranging digits into complete numbers. In our exercise, we analyzed how digits can be placed to form different numbers.

• We started by identifying that we have the digits 3, 2, 3, and 4.
• Recognizing repetitions is vital, as it affects how we generate permutations.
• Next, we accounted for each digit occupying every place in a number (thousands, hundreds, tens, and units).

By systematically assigning digits to each position, we ensure all possible combinations are explored. This step not only helps in identifying unique numbers but also assists in calculating the contributions from each digit correctly in these formed numbers.

Multiset Permutations

Multiset permutations are a particular type of permutation that involves repeated elements. In normal permutations, each element is distinct; however, with multisets, elements may repeat, such as in our set of digits: 3, 2, 3, and 4.When calculating multiset permutations, the formula \[\frac{n!}{p_1!p_2!...p_k!}\] becomes essential. Here, \(n\) is the total number of items, and \(p_1, p_2, ..., p_k\) are the frequencies of the repeating elements. In this exercise:

• \(n = 4\) for the four digits
• Two 3s repeat: \(p_1! = 2!\)

This yields \(\frac{4!}{2!} = 12\) permutations, highlighting another critical aspect of combinatorics. By understanding multiset permutations, we grasp how to manage repeats efficiently to calculate the number of different numbers that can be formed.