The total number of three-digit numbers, with one digit repeated exactly two times, is ______________.

Solution

Maths sir · 4 steps

Do not count the pairs head-on: take all 900 numbers and remove the two cases you do not want.

1. Three kinds only. A three-digit number has either all digits different, exactly one digit repeated twice, or all three digits the same. So the count we want is the total minus the other two kinds.
2. Total and distinct. There are 999 - 100 + 1 = 900 three-digit numbers. With all digits different, the first digit has 9 choices (not 0), the second has 9 (0 is now allowed, but not the first digit), and the third has 8. That gives 9 × 9 × 8 = 648.
3. All digits same. These are 111, 222, …, 999, which makes 9 numbers. The number 000 is not a three-digit number.

   900 - 648 - 9 = 243

4. So. There are 243 three-digit numbers with one digit repeated exactly two times.

The common slip: 252. Subtracting only the 648 numbers with distinct digits gives 252. The 9 numbers like 777 repeat a digit three times, not exactly two, so they must go too.

The idea to keep. When a condition like "exactly two the same" is awkward to count directly, split all cases into types and subtract the types you do not want. Watch the leading zero every time.

Answer from the NTA final answer key. Solution by PrepWiser.