Step-by-step Solution:

Step 1: Analyze the Range
We need natural numbers smaller than \( 2 \times 10^8 \) (which is 200,000,000) using only digits {1, 2}.
This includes all valid 1-digit, 2-digit, ..., up to 8-digit numbers, plus some 9-digit numbers.

Step 2: Count numbers with 1 to 8 digits
For a number with \( n \) digits, each place can be filled by either 1 or 2 (2 choices).
• 1-digit numbers: \( 2^1 = 2 \)
• 2-digit numbers: \( 2^2 = 4 \)
...
• 8-digit numbers: \( 2^8 = 256 \)
Sum of these terms is a Geometric Progression:
\( S_8 = \frac{2(2^8 - 1)}{2 - 1} = 2(255) = 510 \).

Step 3: Count valid 9-digit numbers
The limit is 200,000,000.
A 9-digit number starts with the first digit. Since we can only use {1, 2}:
• If it starts with 2: The number is \( 2... \) which is \( \ge 200,000,000 \). So no numbers starting with 2 are allowed (must be strictly smaller).
• If it starts with 1: The number is \( 1... \) which is definitely \( < 200,000,000 \).
For numbers starting with 1, the remaining 8 spots can be filled with either 1 or 2.
Total 9-digit numbers = \( 1 \times 2^8 = 256 \).

Step 4: Total Count
Total = (1 to 8 digit numbers) + (9 digit numbers starting with 1)
Total = \( 510 + 256 = 766 \).