Calculating Different 5-Digit Numbers

The problem asks us to find the total number of distinct 5-digit numbers that can be formed using the digits 9, 0, 4, 1, and 6, with each digit used exactly once in each number.

A 5-digit number cannot have 0 in the first position (the ten thousand's place). The given digits are 9, 0, 4, 1, 6.

Understanding the Constraints

• We are forming a 5-digit number.
• The digits available are 9, 0, 4, 1, 6.
• Each digit must be used exactly once.
• The first digit (at the ten thousand's place) cannot be 0.

Step-by-Step Calculation Method

Let's consider the positions in the 5-digit number from left to right:

1. First Position (Ten Thousands Place): This position cannot be 0. The available digits are 9, 4, 1, and 6. So, there are 4 choices for the first position.
2. Second Position (Thousands Place): After placing a digit in the first position, we have 4 digits remaining (including 0, as it is now allowed). So, there are 4 choices for the second position.
3. Third Position (Hundreds Place): After placing digits in the first two positions, we have 3 digits remaining. So, there are 3 choices for the third position.
4. Fourth Position (Tens Place): After placing digits in the first three positions, we have 2 digits remaining. So, there are 2 choices for the fourth position.
5. Fifth Position (Units Place): After placing digits in the first four positions, we have 1 digit remaining. So, there is only 1 choice for the fifth position.

To find the total number of different 5-digit numbers, we multiply the number of choices for each position:

Total numbers = (Choices for Pos 1) $\times$ (Choices for Pos 2) $\times$ (Choices for Pos 3) $\times$ (Choices for Pos 4) $\times$ (Choices for Pos 5)

Total numbers = $4 \times 4 \times 3 \times 2 \times 1$

Total numbers = $16 \times 6$

Total numbers = $96$

Alternative Permutation Method

We can also calculate this using the concept of permutations.

The total number of permutations of 5 distinct digits (if there were no restriction on the first digit) is $5!$ (5 factorial).

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

Now, we need to subtract the number of arrangements where 0 is in the first position (which are not valid 5-digit numbers). If 0 is in the first position, we need to arrange the remaining 4 digits (9, 4, 1, 6) in the remaining 4 positions. The number of ways to do this is $4!$ (4 factorial).

$4! = 4 \times 3 \times 2 \times 1 = 24$

The number of valid 5-digit numbers is the total number of permutations minus the number of permutations with 0 in the first position.

Number of 5-digit numbers = Total permutations - Permutations with 0 in the first position

Number of 5-digit numbers = $120 - 24$

Number of 5-digit numbers = $96$

Both methods yield the same result.

Therefore, 96 different 5-digit numbers can be formed from the digits 9, 0, 4, 1, 6.