Let `N` be the number of 4- digit numbers which contain not more than 2 different digits. The sum of the digits of N is :

Options

A. 18
B. 19
C. 20
D. 21

Answer

To solve the problem of finding the number of 4-digit numbers that contain not more than 2 different digits, we can break it down into two cases:

### Step 1: Case 1 - All digits are the same
In this case, all four digits of the number are identical. The valid digits for a 4-digit number range from 1 to 9 (since 0 cannot be the leading digit). Therefore, the possible numbers are:
- 1111, 2222, 3333, ..., 9999

**Calculation**:
- There are 9 choices (1 through 9) for the digit.
- Thus, the total for Case 1 is **9**.

### Step 2: Case 2 - Two different digits
In this case, we can have combinations of two different digits. The first digit cannot be 0 (it must be from 1 to 9), while the second digit can be any digit from 0 to 9, excluding the first digit.

**Step 2.1: Choose the digits**
- Choose the first digit (let's call it A) from 1 to 9. There are 9 choices.
- Choose the second digit (let's call it B) from 0 to 9, excluding A. Therefore, there are 9 choices for B as well (since we can choose from 0 to 9 but not A).

**Step 2.2: Arranging the digits**
Now, we need to arrange these two digits in a 4-digit number. The arrangements can be:
- 3 of A and 1 of B
- 2 of A and 2 of B
- 1 of A and 3 of B

**Calculating arrangements**:
1. **3 A's and 1 B**: The number of arrangements is given by the combination formula:
\[ \text{Arrangements} = \frac{4!}{3!1!} = 4 \]
2. **2 A's and 2 B's**: The number of arrangements is:
\[ \text{Arrangements} = \frac{4!}{2!2!} = 6 \]
3. **1 A and 3 B's**: The number of arrangements is:
\[ \text{Arrangements} = \frac{4!}{1!3!} = 4 \]

**Total arrangements for two different digits**:
- Total arrangements = \( 4 + 6 + 4 = 14 \).

**Step 2.3: Total combinations for Case 2**:
- The total number of 4-digit numbers with two different digits is:
\[ \text{Total for Case 2} = \text{(Choices for A)} \times \text{(Choices for B)} \times \text{(Arrangements)} = 9 \times 9 \times 14 = 1134. \]

### Step 3: Total count of 4-digit numbers
Now, we combine the results from both cases:
\[ N = \text{Case 1} + \text{Case 2} = 9 + 1134 = 1143. \]

### Step 4: Sum of digits of N
Finally, we need to find the sum of the digits of \( N \):
- \( N = 1143 \)
- Sum of digits = \( 1 + 1 + 4 + 3 = 9 \).

### Final Answer
The sum of the digits of \( N \) is **9**.

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