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. a) 18
B. b) 19
C. c) 20
D. d) 21

Answer

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

### Case 1: All four digits are the same
1. A 4-digit number can only have digits from 1 to 9 in the thousands place (the first digit cannot be 0).
2. Therefore, the possible digits that can be used are {1, 2, 3, 4, 5, 6, 7, 8, 9}.
3. Since all four digits are the same, we have 9 possibilities (one for each digit from 1 to 9).

### Case 2: Two different digits are present
1. We will denote the two different digits as A and B.
2. The first digit (thousands place) cannot be 0, so we must choose A from the digits {1, 2, 3, 4, 5, 6, 7, 8, 9} (9 options).
3. The second digit B can be chosen from the digits {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} excluding A. This gives us 9 options for B (since we can choose any digit except A).
4. Now we need to arrange the digits A and B. The arrangements can be calculated as follows:
- The total arrangements of 4 digits where we have 2 A's and 2 B's can be calculated using the formula for permutations of multiset:
\[ \text{Number of arrangements} = \frac{4!}{2! \cdot 2!} = 6 \]
5. Therefore, the total number of 4-digit numbers with two different digits is:
\[ \text{Total for Case 2} = 9 \times 9 \times 6 = 486 \]

### Total Calculation
1. Now, we can find the total number of 4-digit numbers (N):
\[ N = \text{Total from Case 1} + \text{Total from Case 2} = 9 + 486 = 495 \]

### Sum of the digits of N
1. To find the sum of the digits of N (which is 495):
\[ 4 + 9 + 5 = 18 \]

### Conclusion
The sum of the digits of N is **18**.