Problem 95 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The number of odd numbers lying between 40000 and 70000 that can be made from the digits \(0,1,2,4,5,7\) if digits can be repeated in the same number is (A) 864 (B) 932 (C) 766 (D) None of these

Step-by-Step Solution

Answer

The correct option is (D) None of these, as the answer is 1296.

Step 1: Understanding the Constraints

We need to find the number of odd numbers between 40000 and 70000 that can be formed using the digits \( 0, 1, 2, 4, 5, 7 \). The number must be five digits, start with either 4, 5, or 6 (since it needs to be between 40000 and 70000), and end with an odd digit (1, 5, or 7).

Step 2: Choosing the First Digit

The first digit must be 4, 5, or 6 to ensure that the number is between 40000 and 70000. However, the digit 4 and 5 are the only available options since 6 is not part of the available digits. Thus, we have 2 choices for the first digit: 4 or 5.

Step 3: Choosing the Last Digit

Since the number is odd, the last digit must be odd. Therefore, it must be one of \( 1, 5, \) or \( 7 \). So, there are 3 choices for the last digit.

Step 4: Filling the Remaining Digits

The remaining three digits (the second, third, and fourth positions) can be any of the six digits \( 0, 1, 2, 4, 5, 7 \), since digits can be repeated. Thus, for each of these positions, there are 6 choices.

Step 5: Calculating Total Numbers

We have 2 choices for the first digit (4 or 5), 6 choices for each of the second, third and fourth digits, and 3 choices for the last digit (1, 5, or 7). Thus, the total number of combinations can be calculated as: \((2) \times (6^3) \times (3)\).

Step 6: Performing Calculation

Compute the total number of odd numbers: 1. Calculate \(6^3 = 6 \times 6 \times 6 = 216\). 2. Multiply by 2 (for the choices of the first digit): \(216 \times 2 = 432\). 3. Multiply by 3 (for the choices of the last digit): \(432 \times 3 = 1296\).

Key Concepts

Odd Numbers

Odd numbers play a significant role in mathematics. They differ from even numbers as they cannot be divided by 2 without leaving a remainder.
In the context of our exercise, understanding the characteristics of odd numbers is crucial. Odd numbers often end with the digits 1, 3, 5, 7, or 9, but since we are restricted to using the digits 0, 1, 2, 4, 5, 7, our odd numbers can end only with 1, 5, or 7.
This restriction helps us form odd numbers based on the last digit being one of these.

• Numbers have limited options for their final digit to ensure they are odd.
• The selection of 1, 5, or 7 as the last digit plays a strategic role, impacting the count of valid odd numbers formed.

Understanding how the property of being odd restricts certain digits helps to solve the combinatorial problem effectively.

Digit Repetition

Digit repetition allows for more diverse combinations when forming numbers. In this exercise, the ability to repeat digits increases the total number of possible numbers.
Repetition means you can use the same digit in multiple positions without restriction, except for its position in the given rules.

• Repetition allows six choices for each of the second, third, and fourth digits.
• Each choice in digit positions (second to fourth) can be any of the digits 0, 1, 2, 4, 5, or 7.

Without repetition, forming a number that fits all criteria would be significantly more limited. Repetition, therefore, simplifies achieving the required number range while adhering to odd number rules.

Number Range Constraints

Number range constraints determine which digits can be used in specific positions to form valid numbers. These constraints help ensure the number lies within a specified range, like 40000 to 70000.
For this exercise, the range implies specific rules:

• The first digit must ensure the number starts within the given range, so it must be 4 or 5.
• Although 6 could have technically been a viable first digit choice to fall within the range, it is not part of the available digits.

By narrowing down the first digit options, the range constraint simplifies the problem, focusing on finding combinations within the constraints that still allow for odd numbers. This precision is crucial for solving the exercise efficiently.