How many integers in the set {100, 101, 102, ..., 999} have at least one digit repeated?
[This Question was asked as TITA]
[This Question was asked as TITA]
- 252
- 254
- 262
- 264
The Correct Option is A
Approach Solution - 1
We are given the set of integers from 100 to 999, i.e., all 3-digit numbers. To solve the problem, we first calculate the total number of integers in this set.
\[ \text{Total numbers} = 999 - 100 + 1 = 900 \]
Next, we determine how many of these numbers have no repeated digits. Each number has 3 digits: hundreds, tens, and units. Let's evaluate the number of such combinations:
- Hundreds digit: 9 choices (1 through 9, since it cannot be 0)
- Tens digit: 9 choices (excluding the hundreds digit)
- Units digit: 8 choices (excluding the digits used in hundreds and tens)
\[ \text{Numbers with all unique digits} = 9 \times 9 \times 8 = 648 \]
Now, subtract this from the total to find how many numbers have at least one digit repeated:
\[ \text{Repeated digit numbers} = 900 - 648 = \boxed{252} \]
Final Answer:
The number of 3-digit numbers with at least one repeated digit is: \[ \boxed{252} \]
Approach Solution -2
We are to find the number of 3-digit integers between 100 and 999 that contain at least one repeated digit.
Total 3-digit numbers:
The range is from 100 to 999.
So, total 3-digit numbers = \(999 - 100 + 1 = 900\)
Step 1: Count numbers with no repeated digits
We have 10 digits: \(0, 1, 2, ..., 9\)
- The first digit (hundreds place) cannot be 0 → 9 choices (1–9)
- The second digit can be any digit except the first → 9 choices
- The third digit must be different from the first two → 8 choices
So, total 3-digit numbers with all distinct digits: \[ N = 9 \times 9 \times 8 = 648 \]
Step 2: Subtract to find numbers with repeated digits
Total numbers with at least one repeated digit: \[ 900 - 648 = \boxed{252} \]
Final Answer: The number of 3-digit numbers with at least one digit repeated is: \[ \boxed{252} \] Hence, the correct option is (A): \(252\)
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