Problem 91 - A Complete Resource Book in Mathematics for JEE Main 2017

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

39996

Step 1: Understanding the Problem

We need to find the sum of all possible numbers that can be formed from the digits 3, 2, 3, and 4 without repetition. It's helpful to first understand how to arrange these digits and count them efficiently.

Step 2: Calculate Number of Formations

Since the digits are 3, 2, 3, and 4, we have a repetition of the digit '3'. The total number of distinct numbers we can create is given by \(\frac{4!}{2!} = 12\). This accounts for all unique permutations.

Step 3: Find Contribution of Each Digit

Each digit will appear in each place (units, tens, hundreds, thousands) the same number of times across all permutations. Since there are 12 numbers, each digit will appear in each place \(\frac{12}{4} = 3\) times.

Step 4: Calculate Total Contribution

Calculate the sum of digits for one specific place, say units place: - Digit 3 contributes \(3 \times (3 + 3) = 18\),- Digit 2 contributes \(3 \times 2 = 6\),- Digit 4 contributes \(3 \times 4 = 12\).Sum of the units place is \(18 + 6 + 12 = 36\). Since they contribute the same for tens, hundreds, and thousands, multiply 36 by 1111 (1+10+100+1000) to account for all positions.

Step 5: Summing Up All Contributions

Multiplying the sum for one place by 1111 gives the total sum: \(36 \times 1111 = 39996\).

Step 6: Determine the Correct Answer

So the total sum of all numbers formed by the digits is 39996.

Key Concepts

Repetition of Digits

When working with permutations, especially involving digits, repetition can significantly alter the number of possible arrangements. Here, we have the digits 3, 2, 3, and 4, and notice that the digit '3' appears twice. This repetition means we can't simply use the factorial of the number of digits to find the permutations, as it would count some arrangements more than once. Instead, we adjust the permutation calculation using the formula \(\frac{n!}{p!}\), where \(n\) is the total number of digits and \(p\) is the number of times a particular digit repeats. This gives us the correct number of unique permutations. In this problem, \(n = 4\) and \(p = 2\) (since '3' repeats), so the number of unique permutations is \(\frac{4!}{2!} = 12\). This shows how careful accounting for repetition is crucial in many combinatorial problems.

Counting Principle

The counting principle is a fundamental concept used to determine the number of ways that events can occur. It states that if one event can occur in \(m\) ways and a second event can occur independently in \(n\) ways, then the two events can occur together in \(m \times n\) ways. This principle helps when calculating permutations or combinations by providing a structured way to count without needing to enumerate every possibility.In our exercise, the counting principle helps us understand that each digit can occupy each place (units, tens, hundreds, thousands) the same number of times in the total number of permutations. After calculating the permutations, our knowledge that each digit will fill different spots equally ensures balanced contributions from each, as is seen when we calculated that each digit appears an equal number of times in each position of the number.

Digit Contribution in Number Formation

Calculating the contribution of each digit in forming numbers is an insightful approach to solving permutation problems. Remember, in any fixed-length number formed by specific digits, each digit will appear in each place-value position (like units, tens) a certain number of times.Since there are 12 unique permutations, each digit appears in each position \(\frac{12}{4} = 3\) times, for a four-digit number, across all permutations. For instance, in the units place:

• The digit '3' appears three times as either the first or second '3,' contributing a total of \(3 \times 3=9\). Since there are two '3's, it's \(9+9=18\).
• The digit '2' appears three times, contributing \(3 \times 2 = 6\).
• The digit '4' also appears three times, contributing \(3 \times 4 = 12\).

Adding these gives \(18 + 6 + 12 = 36\) for one positional level. To find the total contribution to the number, multiply this positional sum by 1111, reflecting all positions being filled, which results in the grand total contribution to the sum of all numbers formed: \(36 \times 1111 = 39996\). This method allows breaking down complex number formations into simpler, more manageable parts.