Find the total number of six digit numbers `x_(1)x_(2)x_(3)x_(4)x_(5)x_(6)` having the property `x_(1)ltx__(2)lex_(3)ltx_(4)ltx_(5)lex_(6)`.

Answer

To solve the problem of finding the total number of six-digit numbers \( x_1 x_2 x_3 x_4 x_5 x_6 \) 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
We have the following constraints:
- \( x_1 < x_2 \)
- \( x_2 \leq x_3 \)
- \( x_3 < x_4 \)
- \( x_4 < x_5 \)
- \( x_5 \leq x_6 \)

From these constraints, we can see that \( x_1 \) must be the smallest digit and \( x_6 \) must be the largest digit. The digits can range from 0 to 9, but since \( x_1 \) cannot be 0 (as it would make it a five-digit number), \( x_1 \) can take values from 1 to 9.

### Step 2: Identify cases
We can break down the problem into four distinct cases based on the equality constraints:

1. **Case 1**: \( x_1 < x_2 < x_3 < x_4 < x_5 < x_6 \)
2. **Case 2**: \( x_1 < x_2 = x_3 < x_4 < x_5 < x_6 \)
3. **Case 3**: \( x_1 < x_2 < x_3 < x_4 < x_5 = x_6 \)
4. **Case 4**: \( x_1 < x_2 = x_3 < x_4 < x_5 = x_6 \)

### Step 3: Calculate the number of combinations for each case
For each case, we will calculate the number of ways to choose the digits:

- **Case 1**: All digits are distinct. We need to choose 6 different digits from the set {1, 2, ..., 9}. The number of ways to choose 6 digits from 9 is given by \( \binom{9}{6} \).

- **Case 2**: Here, \( x_2 \) and \( x_3 \) are equal. We need to choose 5 distinct digits from {1, 2, ..., 9} (since \( x_1 \) must be less than \( x_2 \)). The number of ways to choose 5 digits from 9 is \( \binom{9}{5} \).

- **Case 3**: In this case, \( x_5 \) and \( x_6 \) are equal. Similar to Case 2, we choose 5 distinct digits from {1, 2, ..., 9}. The number of ways is again \( \binom{9}{5} \).

- **Case 4**: Here, both \( x_2, x_3 \) and \( x_5, x_6 \) are equal. We need to choose 4 distinct digits from {1, 2, ..., 9}. The number of ways is \( \binom{9}{4} \).

### Step 4: Add the cases together
Now we sum the combinations from all four cases:

\[ \text{Total} = \binom{9}{6} + \binom{9}{5} + \binom{9}{5} + \binom{9}{4} \]

### Step 5: Simplify using combination properties
Using the property of combinations, we can simplify the expression:

\[ \binom{9}{5} + \binom{9}{5} = 2 \cdot \binom{9}{5} \]

Thus, we can rewrite the total as:

\[ \text{Total} = \binom{9}{6} + 2 \cdot \binom{9}{5} + \binom{9}{4} \]

Using the identity \( \binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r} \), we can combine terms:

\[ = \binom{10}{6} + \binom{10}{5} \]

Finally, we can use the identity again:

\[ = \binom{11}{6} \]

### Final Answer
Thus, the total number of six-digit numbers \( x_1 x_2 x_3 x_4 x_5 x_6 \) satisfying the given conditions is:

\[ \binom{11}{6} \]