The number of integers n with `100 le n le 999` and containing at most two distinct digits is

Options

A. 252
B. 280
C. 324
D. 360

Answer

To find the number of integers \( n \) such that \( 100 \leq n \leq 999 \) and containing at most two distinct digits, we can break down the problem into several cases based on the number of distinct digits.

### Step 1: Count numbers with only one distinct digit
These numbers will be of the form \( aaa \), where \( a \) is a digit from 1 to 9 (since \( n \) must be a three-digit number). The valid numbers are:
- 111
- 222
- 333
- 444
- 555
- 666
- 777
- 888
- 999

Thus, there are **9** such numbers.

### Step 2: Count numbers with exactly two distinct digits
For numbers with exactly two distinct digits, we can have combinations of the digits in different positions. Let's denote the two distinct digits as \( a \) and \( b \).

#### Case 2.1: The first digit is \( a \) and the second digit is \( b \)
- The first digit \( a \) can be any digit from 1 to 9 (9 choices).
- The second digit \( b \) can be any digit from 0 to 9, but it must be different from \( a \) (9 choices).

Now we can form numbers in the following patterns:
1. \( aab \)
2. \( aba \)
3. \( baa \)
4. \( abb \)
5. \( bab \)
6. \( bba \)

For each combination of \( a \) and \( b \), we can form **6** different numbers. Therefore, the total for this case is:
\[ 9 \text{ (choices for } a\text{)} \times 9 \text{ (choices for } b\text{)} \times 6 \text{ (arrangements)} = 486 \]

### Step 3: Total Count
Now, we add the results from Step 1 and Step 2:
\[ \text{Total} = 9 \text{ (one distinct digit)} + 486 \text{ (two distinct digits)} = 495 \]

Thus, the total number of integers \( n \) such that \( 100 \leq n \leq 999 \) and containing at most two distinct digits is **495**.