Problem 117 - A Complete Resource Book in Mathematics for JEE Main

Question

The number of numbers greater than \(10^{6}\) that can be formed using the digits of the number 2334203, if all the digits of the given number must be used, is (A) 360 (B) 420 (C) 260 (D) None of these

Step-by-Step Solution

Answer

(A) 360

Step 1: Analyze the Digits

First, identify the digits in the number 2334203. They are: 2, 3, 3, 4, 2, 0, and 3. Notice that there are 7 digits in total. To form a number greater than \(10^6\), we need a 7-digit number, which means all digits must be used.

Step 2: Condition for Greater Than \(10^6\)

Any 7-digit number will naturally be greater than \(10^6\) since it has more digits. However, there's a 0 among the digits, and a valid number cannot start with 0. Therefore, we need to ensure that numbers do not start with 0.

Step 3: Calculate Total Permutations

Calculate the total permutations of the digits 2334203 without regard to starting digit. The formula for permutations with repeated elements is: \[ \frac{7!}{2! \, 3!} \]. Here, 2! is for the two '2's, and 3! is for the three '3's. Calculating, we get: \( \frac{7!}{2! \, 3!} = \frac{5040}{2 \times 6} = 420 \).

Step 4: Subtract Invalid Cases Starting with Zero

Now calculate invalid permutations starting with 0. If 0 is to be the leading digit, we consider the digits 2, 3, 3, 4, 2, 3 for the remaining 6 positions. The permutations are: \( \frac{6!}{2! \, 3!} = \frac{720}{2 \times 6} = 60 \).

Step 5: Calculate Valid Numbers

Subtract the invalid permutations (starting with 0) from the total permutations to get the valid numbers: \( 420 - 60 = 360 \).

Step 6: Verify the Approach

Review the steps and ensure the math was correctly applied at each stage. Ensure the subtraction of the 60 invalid cases aligns with the understanding that any number starting with 0 is invalid.

Key Concepts

Permutations with Repetition

Permutations concern the different ways we can arrange a set of items. When dealing with permutations with repetition, we're interested in arrangements where some items may repeat. This is especially relevant in scenarios involving identical elements.
For example, in the number 2334203, certain digits are repeated. This affects how we calculate the total number of permutations since switching identical digits does not create a new arrangement.

• Identify the total number of items to arrange (7 in this case).
• Identify the repeating items (e.g., two '2's and three '3's). Their repetitions must be accounted for using factorial, denoted as 2! for the two '2’s and 3! for the three '3’s.
• Use the formula for permutations with repetition: \[ \frac{n!}{p_1! \, p_2! \ldots \ p_k!} \]where \( n \) is the total items, and \( p_1, p_2, \ldots\) are the counts of identical items.

Understanding permutations with repetition ensures that each arrangement is counted correctly without oversimplification.

Factorial Calculations

Factorials are critical in calculating permutations, especially when repetition is involved. The factorial of a number \( n \), written as \( n! \), represents the product of all integers from 1 to \( n \). For illustration: \[ n! = n \times (n-1) \times (n-2) \times \ldots \times 2 \times 1 \]This is straightforward for small numbers but grows quickly as \( n \) increases.
Factorial calculations help in determining the total possible arrangements of a set of items. They simplify complex arithmetic tasks, especially when casting arrangements that include repeated elements:

• The factorial of 7, denoted as 7!, is \( 7! = 5040 \).
• For repeated items, compute the factorial for each group: \( 2! = 2 \) for the two '2's, and \( 3! = 6 \) for the three '3's.
• These factorials are used to adjust the total arrangement count, preventing overcounting of identical items.

Mastering factorial calculations ensures that you're accurately handling permutations and combinations.

Number Properties

Number properties are essential in understanding what constitutes valid numbers in permutations, especially when constraints like 'the number must be greater than a million' are in play. Recognizing specific intrinsic properties can make a big difference:
To form numbers greater than \(10^6\), the number must utilize all its digits (in this example, comprising seven digits). Additionally, any number using all seven given digits will inherently exceed a million.

• Digits must be arranged to avoid beginning with zero, as numbers like "0233423" are not valid.
• Recognizing this rule helps in determining which permutations to exclude.
• Ensuring digits adhere to these positional requirements prevents the formation of invalid numbers and maintains accuracy in permutation calculations.

Understanding these properties clarifies the need for excluding certain permutations to achieve accurate results.

Mathematics Problem-Solving

Problem-solving in mathematics involves a structured method to break down and solve tasks. Using approaches like breaking down the problem into smaller steps can enhance understanding and effectiveness:
1. **Identify Key Elements**: Begin by understanding what digits or elements you have, and count repetitions.
2. **Understand Constraints**: Grasp any conditions the problem sets, such as number types or arrangement rules.
3. **Execute the Methodology**: Apply the relevant formulas and mathematical principles, such as the appropriate permutation formula, considering repetition and factorial calculations.

• Calculate total permutations using the full set of digits: \( \frac{7!}{2! \, 3!} = 420 \).
• Assess the invalid cases (leading zeros): \( \frac{6!}{2! \, 3!} = 60 \).
• Subtract invalid cases from the total to arrive at the final count.

Through this strategic positioning, you build toward reliable solutions with confidence, refining both your accuracy and efficiency in problem-solving.