Problem 4 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the number of five digit numbers with distinct digits and 2 at the \(10^{\text {a }}\) place is \(336 \mathrm{k}\), then \(\mathrm{k}\) is equal to: (a) 4 (b) 6 (c) 7 (d) 8

Step-by-Step Solution

Answer

There seems to be a misalignment with the problem statement options; the calculated k is 9.

Step 1: Understand the Position Fixed

The problem specifies that we need a five-digit number where the digit '2' is fixed in the ten-thousands place. Let's denote a five-digit number as ABCDE, where A is fixed to '2'.

Step 2: Determine the Remaining Digits

After fixing '2' at the ten-thousands place, we have four places left (BCDE) and nine digits (0, 1, 3, 4, 5, 6, 7, 8, 9) available.

Step 3: Count Permutations of Remaining Digits

The remaining four digits need to be filled using the available 9 digits, and each digit must be distinct. The number of ways to select and arrange 4 digits from 9 is calculated by: \[ 9 imes 8 imes 7 imes 6 \] This accounts for selecting a digit for each remaining place (thousands, hundreds, tens, and units).

Step 4: Calculate Total Numbers and Relation to k

Calculate the total number of five-digit numbers by computing the permutation: \[ 9 imes 8 imes 7 imes 6 = 3024 \]According to the problem, this total is given as \(336k\), so we have the equation: \[ 336k = 3024 \]

Step 5: Solve for k

To find \(k\), divide the total number 3024 by 336:\[ k = \frac{3024}{336} = 9 \] Given the options, it appears there is a mismatch in expected outcomes due to a constraint or condition interpretation.

Key Concepts

Five-Digit Numbers

A five-digit number is simply a number that has exactly five digits. These numbers range from 10,000 to 99,999. In this exercise, we are particularly interested in forming five-digit numbers with specific characteristics, like having distinct digits, which adds an interesting twist to our counting problem.

To form a five-digit number, imagine each digit as a position in a sequence. These positions are labeled as ten-thousands, thousands, hundreds, tens, and units. The goal is to fill each position with a digit between 0 and 9 while adhering to any constraints laid out by the problem, such as having the digit '2' fixed in one specific place.

A major consideration in this exercise is how limiting it is to demand distinct digits. This constraint significantly affects the number of possibilities and is a central focus of permutation problems in competitive exams like JEE Main Mathematics.

Distinct Digits

In this context, 'distinct digits' means that each digit in the number must be unique and cannot be repeated. Distinct digits are essential in permutation problems as they restrict arrangements and reduce the total possibilities.

When you are arranging a number with distinct digits, it's crucial to understand how to count the permutations properly. This requires selecting digits in a specific order without repetition, which is typically calculated using a series of multiplying decreasing numbers. For example, if you choose from 9 digits, as the problem specifies, you begin by picking any one of the 9, then 8 options remain for the next, and so on.

• This means for four digits, we calculate it as: \( 9 \times 8 \times 7 \times 6 \).

This concept is frequently tested in competitive exams, illustrating the importance of correctly understanding permutations with distinct digits.

Ten-Thousands Place

The ten-thousands place is the leftmost position in any five-digit number and is crucial because it dictates the scale or size of the number. In this exercise, the digit '2' is fixed in this position, fundamentally shaping the rest of the number formation process.

Fixing a digit in the ten-thousands place reduces the number of slots you need to fill, leaving us with four more places (the remaining digits) to solve for. This impacts the way we count the possible permutations of the number.

• This task asks us to consider which digits are left for the thousands, hundreds, tens, and unit places.
• Here, out of the original 10 digits (0 through 9), only 9 choices remain since '2' is already used.

Grasping how fixing a number in the ten-thousands place alters your counting will aid in solving permutation problems efficiently.

JEE Main Mathematics

The JEE Main Mathematics exam is a highly competitive and challenging test taken by students aspiring for engineering colleges in India. Permutation problems, like the one in this exercise, are a staple in the examination because they test students' understanding of fundamental counting principles.

This type of exercise not only tests permutations but also applies additional constraints. Understanding how to manipulate permutations under constraints is a critical skill for the JEE Main Exam.

• Key concepts include properly fixing known digits in positions.
• Understanding available options for remaining digits.
• Calculating permutations under constraints effectively.

Mastering these concepts through practice problems like this ensures preparedness for entering the competitive world of engineering entrance exams.