Understanding 5-Digit Number Formation with Fixed Digits

The problem asks us to find the number of different 5-digit numbers that can be formed using the digits 9, 0, 4, 1, and 6, with the specific condition that the digit '0' must be placed in the tenth place.

The available digits are {9, 0, 4, 1, 6}. These are 5 distinct digits.

A 5-digit number has five places:

• Ten Thousands place (leftmost digit)
• Thousands place
• Hundreds place
• Tens place
• Units place (rightmost digit)

The constraint is that the digit '0' is in the "tenth place". In the context of number places, "tens place" refers to the second digit from the right (e.g., in 12345, 4 is in the tens place). Assuming "tenth place" is a slightly unusual phrasing for "tens place", let's proceed with this interpretation.

So, the 5-digit number will have the structure:

\(\_ \_ \_ \text{0} \_\)

Where the '0' is fixed in the Tens place.

The digits we have left to fill the remaining four places (Ten Thousands, Thousands, Hundreds, and Units) are {9, 4, 1, 6}. These are 4 distinct digits.

Let's fill the places from left to right:

• The Ten Thousands place must be filled by one of the remaining digits {9, 4, 1, 6}. Since none of these digits is 0, any of them can be placed in the Ten Thousands place. There are 4 choices for this place.
• The Thousands place must be filled by one of the remaining 3 digits (after one digit is used for the Ten Thousands place). There are 3 choices for this place.
• The Hundreds place must be filled by one of the remaining 2 digits. There are 2 choices for this place.
• The Units place must be filled by the last remaining digit. There is 1 choice for this place.

Alternatively, we have 4 distinct digits {9, 4, 1, 6} to arrange in the 4 remaining places. The number of ways to arrange \(n\) distinct items in \(n\) places is given by \(n!\) (n factorial).

In this case, we need to arrange 4 distinct digits in 4 places. The number of arrangements is \(4!\).

So, there are 24 different ways to arrange the remaining digits {9, 4, 1, 6} in the remaining four places, while keeping '0' fixed in the tens place.

Therefore, the total number of different 5-digit numbers that can be formed is 24.

If we interpreted "tenth place" as the second digit from the left (Thousands place), the number structure would be \(\_ \text{0} \_ \_ \_\). The remaining digits {9, 4, 1, 6} fill the 1st, 3rd, 4th, and 5th places. The first place can be filled by any of {9, 4, 1, 6} (4 choices, as none is 0). The remaining 3 places are filled by the remaining 3 digits in \(3! = 6\) ways. Total numbers would be \(4 \times 6 = 24\). Both plausible interpretations lead to the same result.

Calculating Permutations for Number Formation

To form the 5-digit number \(\_ \_ \_ \text{0} \_\), we fix the digit '0' in the tens place.

We use the remaining digits {9, 4, 1, 6} to fill the other four positions.

The positions to fill are:

• Ten Thousands place
• Thousands place
• Hundreds place
• Units place

Since the available digits for these positions are {9, 4, 1, 6}, which are all non-zero, the first position (Ten Thousands place) can be any of these 4 digits.

The number of ways to arrange these 4 distinct digits in the 4 remaining places is the number of permutations of 4 items taken 4 at a time, denoted as \(P(4,4)\) or \(4!\).

Thus, there are 24 different 5-digit numbers possible under the given conditions.