The number of numbers, lying between 99 and 1000 that can be made from the digits 2, 3, 7, 0, 8 and 6 when the digits occur only once in each number, is

Options

A. 100
B. 90
C. 120
D. 80

Answer

To solve the problem of finding the number of numbers that can be formed using the digits 2, 3, 7, 0, 8, and 6, which lie between 99 and 1000, we will follow these steps:

### Step 1: Identify the range of numbers
We need to find 3-digit numbers since the numbers must lie between 99 and 1000.

### Step 2: Determine the available digits
The available digits are: 2, 3, 7, 0, 8, and 6. This gives us a total of 6 digits.

### Step 3: Establish the first digit
The first digit of a 3-digit number cannot be 0 (as it would not be a 3-digit number). Therefore, the possible choices for the first digit are 2, 3, 7, 8, and 6. This gives us 5 options for the first digit.

### Step 4: Determine the second digit
The second digit can be any of the 6 digits (2, 3, 7, 0, 8, 6) since we can use 0 here. However, we must ensure that the second digit is different from the first digit chosen. Thus, after choosing the first digit, we have 5 remaining choices for the second digit.

### Step 5: Determine the third digit
The third digit can also be any of the 6 digits, but it must be different from both the first and second digits chosen. Hence, after selecting the first and second digits, we will have 4 choices left for the third digit.

### Step 6: Calculate the total number of combinations
Now, we can calculate the total number of 3-digit numbers that can be formed:

\[ \text{Total combinations} = (\text{Choices for 1st digit}) \times (\text{Choices for 2nd digit}) \times (\text{Choices for 3rd digit}) \]

Substituting the values we found:

\[ \text{Total combinations} = 5 \times 5 \times 4 \]

Calculating this gives:

\[ 5 \times 5 = 25 \]
\[ 25 \times 4 = 100 \]

### Final Answer
Thus, the total number of numbers that can be formed between 99 and 1000 using the given digits is **100**.