Final answer:
To find the number of natural numbers between 100 and 1000 that have at least one digit as 7, we can break down the problem into cases. The correct answer is none of the given options (a), (b), or (c), but 990. Therefore, the correct answer is (d) None.
Step-by-step explanation:
To find the number of natural numbers between 100 and 1000 that have at least one digit as 7, we can break down the problem into cases.
Case 1: Numbers with one digit as 7:
We have 9 choices for the hundreds place (1 to 9) and 10 choices each for the tens and units places (0 to 9).
So, the number of numbers with one digit as 7 is 9 x 10 x 10 = 900.
Case 2: Numbers with two digits as 7:
We have 9 choices for the hundreds place (1 to 9) and 1 choice for the units place (7).
We can choose any of the remaining 10 digits (0 to 9) for the tens place.
So, the number of numbers with two digits as 7 is 9 x 1 x 10 = 90.
Case 3: Numbers with three digits as 7:
We have 1 choice for the hundreds place (7) and 10 choices each for the tens and units places (0 to 9).
So, the number of numbers with three digits as 7 is 1 x 10 x 10 = 100.
Total: Adding the numbers from all three cases, we get 900 + 90 + 100 = 990.
Therefore, the correct answer is (d) None, since there are 990 natural numbers between 100 and 1000 that have at least one digit as 7, not any of the given options (a), (b), or (c).