Understanding the Problem: Forming 5-Digit Numbers

The question asks us to consider 5-digit numbers formed using a specific set of digits: 0, 1, 2, 4, and 5. The key constraint is that each digit can be used only once (without repetition).

We need to figure out two things:

1. The total number of distinct 5-digit numbers that can be formed using these digits without repetition.
2. The number of these 5-digit numbers that are strictly greater than 50,000.

Finally, we will calculate the percentage of numbers greater than 50,000 out of the total number of 5-digit numbers formed.

Calculating the Total Number of 5-Digit Numbers

We have 5 distinct digits: 0, 1, 2, 4, 5. We want to form a 5-digit number.

A 5-digit number has 5 places:

_ _ _ _ _

The first digit of a 5-digit number cannot be 0. So, for the first place (the ten thousands place), we have a limited choice.

• First digit: Cannot be 0. The available digits are 1, 2, 4, or 5. There are 4 choices for the first digit.
• Second digit: After choosing the first digit, we have 4 digits remaining (including 0, which is now allowed in the second place). So, there are 4 choices for the second digit.
• Third digit: We have used two digits, so 3 digits remain. There are 3 choices for the third digit.
• Fourth digit: We have used three digits, so 2 digits remain. There are 2 choices for the fourth digit.
• Fifth digit: Only 1 digit remains. There is 1 choice for the fifth digit.

The total number of distinct 5-digit numbers that can be formed is the product of the number of choices for each position:

Total numbers = (Choices for 1st digit) $\times$ (Choices for 2nd digit) $\times$ (Choices for 3rd digit) $\times$ (Choices for 4th digit) $\times$ (Choices for 5th digit)

Total numbers = $4 \times 4 \times 3 \times 2 \times 1 = 4 \times 4! = 4 \times 24 = 96$.

So, there are 96 unique 5-digit numbers possible using these digits without repetition.

Finding Numbers Greater Than 50,000

A 5-digit number formed using these digits will be greater than 50,000 if and only if its first digit (the ten thousands place) is 5.

Let's look at the 5 places again:

_ _ _ _ _

• First digit: Must be 5. There is only 1 choice for the first digit.

Now, we have used the digit 5. The remaining available digits are 0, 1, 2, and 4. We need to arrange these 4 remaining digits in the remaining 4 places (second, third, fourth, and fifth).

• Second digit: Any of the remaining 4 digits (0, 1, 2, 4). There are 4 choices.
• Third digit: Any of the remaining 3 digits. There are 3 choices.
• Fourth digit: Any of the remaining 2 digits. There are 2 choices.
• Fifth digit: The last remaining digit. There is 1 choice.

The number of distinct 5-digit numbers greater than 50,000 is the product of the number of choices for each position:

Numbers > 50,000 = (Choices for 1st digit) $\times$ (Choices for 2nd digit) $\times$ (Choices for 3rd digit) $\times$ (Choices for 4th digit) $\times$ (Choices for 5th digit)

Numbers > 50,000 = $1 \times 4 \times 3 \times 2 \times 1 = 1 \times 4! = 1 \times 24 = 24$.

So, there are 24 unique 5-digit numbers greater than 50,000 formed using these digits without repetition.

Calculating the Percentage

To find the percentage of numbers which are greater than 50,000, we use the formula:

Percentage = $\left( \frac{\text{Number of numbers greater than 50,000}}{\text{Total number of 5-digit numbers}} \right) \times 100\%$

Percentage = $\left( \frac{24}{96} \right) \times 100\%$

Simplify the fraction:

$\frac{24}{96} = \frac{1}{4}$

Now, calculate the percentage:

Percentage = $\frac{1}{4} \times 100\% = 25\%$

Conclusion

Out of the 96 distinct 5-digit numbers that can be formed using the digits 0, 1, 2, 4, and 5 without repetition, 24 of them are greater than 50,000. This represents 25% of the total numbers.

Revision Table: 5-Digit Number Formation

Additional Information: Permutations and Combinations Basics

This problem primarily involves the concept of permutations, which is about arranging items where the order matters. When forming numbers, the order of the digits is crucial (e.g., 123 is different from 321).

• Permutations ($nP_r$): The number of ways to arrange $r$ objects from a set of $n$ distinct objects, where order matters. The formula is $nP_r = \frac{n!}{(n-r)!}$. In our problem, for the remaining digits after placing the first one, we arrange 4 digits out of 4, which is $4P_4 = \frac{4!}{(4-4)!} = \frac{4!}{0!} = 4! = 24$.
• Factorial ($n!$): The product of all positive integers up to $n$. $n! = n \times (n-1) \times \dots \times 2 \times 1$. $0!$ is defined as 1.
• Combinations ($nC_r$): The number of ways to choose $r$ objects from a set of $n$ distinct objects, where order does not matter. The formula is $nC_r = \frac{n!}{r!(n-r)!}$. Combinations are not directly used in this problem because the arrangement of digits matters.

Understanding the difference between permutations and combinations is key in solving problems involving counting arrangements or selections.