To solve the problem of finding the number of six-digit numbers that can be formed using the digits 1, 2, 3, 4, 5, and 6, with the condition that not more than two digits smaller than a given digit can appear to its right, we can follow these steps:
### Step-by-Step Solution:
1. **Understanding the Condition**:
- For each digit in the number, we need to ensure that there are not more than two digits smaller than that digit to its right. This means that for the largest digit (6), it can only be placed in the first three positions of the number.
2. **Placement of the Digit 6**:
- The digit 6 can occupy any of the first three positions (1st, 2nd, or 3rd). This gives us 3 choices for placing the digit 6.
3. **Placement of the Digit 5**:
- If 6 is placed in one of the first three positions, the digit 5 can also occupy any of the first three positions (but not the position already occupied by 6). Therefore, 5 also has 3 choices for its placement.
4. **Placement of the Digit 4**:
- Following the same reasoning, the digit 4 can also occupy any of the first three positions, giving it 3 choices as well.
5. **Placement of the Digits 3, 2, and 1**:
- After placing 6, 5, and 4, we will have three positions filled. The remaining three digits (3, 2, and 1) can be placed in the remaining three positions. There are 3! (which is 6) ways to arrange these three digits.
6. **Calculating the Total Arrangements**:
- Since each of the digits 6, 5, and 4 has 3 choices, and the arrangement of the remaining digits can be done in 6 ways, we can calculate the total number of valid arrangements as follows:
\[
\text{Total arrangements} = 3 \times 3 \times 3 \times 3! = 3^4 \times 6
\]
7. **Final Calculation**:
- Calculate \(3^4\):
\[
3^4 = 81
\]
- Now multiply by 6:
\[
81 \times 6 = 486
\]
Thus, the total number of six-digit numbers that can be formed under the given conditions is **486**.
