How many numbers each lying between 100 and 1000 can be formed with the digits 0,2,3,4,5 ( no digit being repeated ) ?

Options

A. 24
B. 48
C. 72
D. 96

Answer

To solve the problem of how many numbers between 100 and 1000 can be formed using the digits 0, 2, 3, 4, and 5 without repeating any digits, we can follow these steps:

### Step 1: Identify the range and constraints
We need to form three-digit numbers that are greater than 100 and less than 1000. This means the first digit (hundreds place) cannot be 0.

### Step 2: Determine the choices for the first digit
The first digit can be any of the digits 2, 3, 4, or 5 (since 0 cannot be used). Therefore, we have 4 options for the first digit.

**Choices for the first digit:** 2, 3, 4, 5
**Total choices for the first digit:** 4

### Step 3: Determine the choices for the second digit
After selecting the first digit, we cannot use that digit again. We also can use 0 in the second position. Therefore, the second digit can be chosen from the remaining 4 digits (including 0).

**Choices for the second digit:** 0 and the three remaining digits
**Total choices for the second digit:** 4

### Step 4: Determine the choices for the third digit
After selecting the first and second digits, we will have 3 digits left to choose from for the third digit.

**Choices for the third digit:** Remaining 3 digits
**Total choices for the third digit:** 3

### Step 5: Calculate the total combinations
To find the total number of three-digit combinations, we multiply the number of choices for each digit:

\[ \text{Total combinations} = (\text{Choices for 1st digit}) \times (\text{Choices for 2nd digit}) \times (\text{Choices for 3rd digit}) = 4 \times 4 \times 3 \]

Calculating this gives:

\[ 4 \times 4 = 16 \\ 16 \times 3 = 48 \]

### Final Answer
Thus, the total number of three-digit numbers that can be formed is **48**.

---