To solve the problem of finding the total number of six-digit numbers \( x_1x_2x_3x_4x_5x_6 \) such that \( x_1 < x_2 \leq x_3 < x_4 < x_5 \leq x_6 \), we can follow these steps:
### Step 1: Understand the constraints
The digits \( x_1, x_2, x_3, x_4, x_5, x_6 \) must be chosen from the digits 1 to 9 (since 0 cannot be the leading digit of a six-digit number). The constraints imply that:
- \( x_1 < x_2 \)
- \( x_2 \leq x_3 \)
- \( x_3 < x_4 \)
- \( x_4 < x_5 \)
- \( x_5 \leq x_6 \)
### Step 2: Transform the problem
To simplify the counting, we can redefine the variables to account for the equalities:
- Let \( y_1 = x_1 \)
- Let \( y_2 = x_2 \)
- Let \( y_3 = x_3 \)
- Let \( y_4 = x_4 \)
- Let \( y_5 = x_5 \)
- Let \( y_6 = x_6 \)
We can express the conditions as:
- \( y_1 < y_2 \)
- \( y_2 \leq y_3 \)
- \( y_3 < y_4 \)
- \( y_4 < y_5 \)
- \( y_5 \leq y_6 \)
### Step 3: Count the distinct digits
To satisfy the inequalities, we can think of choosing 6 digits from the set {1, 2, 3, 4, 5, 6, 7, 8, 9}. Since we have 9 possible digits and we need to choose 6, we can use combinations.
### Step 4: Use combinations
We can break this down into cases based on the number of distinct digits selected. The maximum number of distinct digits we can choose is 9, and we need to account for the cases where some digits can be repeated due to the inequalities.
1. **Case 1**: All digits are distinct. This corresponds to choosing 6 different digits from 9, which is \( \binom{9}{6} \).
2. **Case 2**: One digit is repeated. This corresponds to choosing 5 different digits from 9, which is \( \binom{9}{5} \).
3. **Case 3**: Two digits are repeated. This corresponds to choosing 4 different digits from 9, which is \( \binom{9}{4} \).
### Step 5: Calculate the total
Now we can sum the combinations from all cases:
\[
\text{Total} = \binom{9}{6} + \binom{9}{5} + \binom{9}{4}
\]
Using the combination formula \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \):
- \( \binom{9}{6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \)
- \( \binom{9}{5} = \binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \)
- \( \binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \)
Thus, the total is:
\[
\text{Total} = 84 + 126 + 126 = 336
\]
### Final Answer
The total number of six-digit numbers \( x_1x_2x_3x_4x_5x_6 \) satisfying the given conditions is **336**.
