How many 3-digit numbers are there in between 100 and 300, having first and the last digit as 2?

### Concept & Logic This is a permutations problem with fixed constraints. When specific digits of a number are fixed, you only need to determine the number of possible variations for the remaining unconstrained digits to find the total count. ### Step-by-Step Solution * **Given:** We need $3$-digit numbers between $100$ and $300$. The first digit must be $2$, and the last digit must be $2$. * **Step 1: Set up the number structure.** Since it is a $3$-digit number starting and ending with $2$, it must take the format: **$2 \_ 2$** * **Step 2: Determine the possibilities for the middle digit.** The middle digit is the tens place. Since there are no restrictions placed on this middle digit, it can be any valid single digit from $0$ to $9$. The possible digits are: $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$. * **Step 3: Count the valid numbers.** Since there are exactly $10$ possible digits that can fill the middle blank, there are $10$ possible numbers. The numbers are: $202, 212, 222, 232, 242, 252, 262, 272, 282, 292$. ### Exam Strategy & Shortcut Visualize the format immediately: `[Fixed] [Variable] [Fixed]`. The number is bounded strictly by the middle digit. Since a base-$10$ digit has exactly $10$ possibilities ($0$ through $9$), the answer is inherently $10$. No listing or manual counting is required. ### Common Pitfall Students sometimes overthink the "between $100$ and $300$" constraint and wonder if they need to exclude boundaries, or they mistakenly exclude the digit $2$ from the middle spot assuming digits cannot be repeated (which wasn't stated in the problem). Always assume digits can repeat unless explicitly forbidden. ### Final Answer **Therefore, the correct answer is 10.**