Calculating the Number of 6-Digit Numbers

The question asks us to find out how many different 6-digit numbers can be formed using the digits 4, 5, 2, 1, 8, and 9. We have a set of 6 distinct digits: {1, 2, 4, 5, 8, 9}. We need to arrange all 6 of these digits to form a 6-digit number.

Since the order of the digits matters when forming a number (for example, 452189 is a different number than 981254), this is a problem of permutation.

Understanding Permutations

A permutation is an arrangement of objects in a specific order. When we arrange all the items from a set of $n$ distinct items, the number of possible permutations is given by the factorial of $n$, denoted as $n!$.

The formula for the number of permutations of $n$ distinct objects taken all at a time is:

\(P(n, n) = n!\)

where \(n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1\).

Applying Permutations to Form 6-Digit Numbers

In this specific problem, we have 6 distinct digits (4, 5, 2, 1, 8, 9) and we need to form a 6-digit number using all of them. This means we are arranging 6 distinct items in 6 positions.

Here, the number of distinct digits is \(n = 6\).

The number of different 6-digit numbers that can be formed is the number of permutations of 6 distinct digits taken all at a time, which is \(6!\).

Calculating the Number of Different 6-Digit Numbers

Let's calculate the value of \(6!\):

\(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1\)

Step-by-step calculation:

• \(6 \times 5 = 30\)
• \(30 \times 4 = 120\)
• \(120 \times 3 = 360\)
• \(360 \times 2 = 720\)
• \(720 \times 1 = 720\)

So, the number of different 6-digit numbers that can be formed from the digits 4, 5, 2, 1, 8, 9 is 720.

Conclusion

Using the concept of permutations, we found that there are 720 different ways to arrange the 6 distinct digits (4, 5, 2, 1, 8, 9) to form unique 6-digit numbers.