Step 1: Understanding the Concept:
We need to count 3-digit numbers \( xyz \). The conditions are \( x<y \) and \( z<y \).
Also, since it's a 3-digit number, \( x \in \{1, 2, \dots, 9\} \).
The digit \( y \) must be greater than \( x \), so \( y \in \{1, 2, \dots, 9\} \). However, since \( x \ge 1 \), \( y \) must be at least 2.
The digit \( z \) can be any digit from \( \{0, 1, \dots, 9\} \) as long as it is less than \( y \).
Step 2: Key Formula or Approach:
We will use summation by fixing the value of the middle digit \( y \). For each possible value of \( y \), we calculate the number of choices for \( x \) and \( z \).
- Choices for \( x \): Number of integers in \( \{1, 2, \dots, y-1\} \) is \( y-1 \).
- Choices for \( z \): Number of integers in \( \{0, 1, \dots, y-1\} \) is \( y \).
Total ways for a fixed \( y \) is \( (y-1) \times y \).
Step 3: Detailed Explanation:
The possible values for \( y \) are from 2 to 9 (if \( y=1 \), then \( x<1 \implies x=0 \), which is not allowed).
Let's sum the products \( (y-1)y \) for \( y = 2, 3, \dots, 9 \):
- If \( y = 2 \): \( x \in \{1\} \), \( z \in \{0, 1\} \). Total \( = 1 \times 2 = 2 \).
- If \( y = 3 \): \( x \in \{1, 2\} \), \( z \in \{0, 1, 2\} \). Total \( = 2 \times 3 = 6 \).
- If \( y = 4 \): \( x \in \{1, 2, 3\} \), \( z \in \{0, 1, 2, 3\} \). Total \( = 3 \times 4 = 12 \).
- If \( y = 5 \): \( x \in \{1, 2, 3, 4\} \), \( z \in \{0, 1, 2, 3, 4\} \). Total \( = 4 \times 5 = 20 \).
- If \( y = 6 \): \( x \in \{1, \dots, 5\} \), \( z \in \{0, \dots, 5\} \). Total \( = 5 \times 6 = 30 \).
- If \( y = 7 \): \( x \in \{1, \dots, 6\} \), \( z \in \{0, \dots, 6\} \). Total \( = 6 \times 7 = 42 \).
- If \( y = 8 \): \( x \in \{1, \dots, 7\} \), \( z \in \{0, \dots, 7\} \). Total \( = 7 \times 8 = 56 \).
- If \( y = 9 \): \( x \in \{1, \dots, 8\} \), \( z \in \{0, \dots, 8\} \). Total \( = 8 \times 9 = 72 \).
Summing these up:
\[ \text{Total} = 2 + 6 + 12 + 20 + 30 + 42 + 56 + 72 \]
\[ \text{Total} = (2+6) + (12+20) + (30+42) + (56+72) \]
\[ \text{Total} = 8 + 32 + 72 + 128 \]
\[ \text{Total} = 40 + 200 = 240 \]
Alternative calculation using summation formula \( \sum_{y=2}^{9} (y^2 - y) \):
\[ \sum_{y=1}^{9} y^2 - \sum_{y=1}^{9} y - (1^2 - 1) \]
\[ = \frac{9(10)(19)}{6} - \frac{9(10)}{2} = 285 - 45 = 240 \]
Step 4: Final Answer:
The number of such 3-digit numbers is 240.