The sequence is "xxx333xx". The third digit cannot be 3, so it must be 4 (1 choice). The seventh digit must be 4, and the last digit can be either 3 or 4, giving \(1 \times 2 = 2\) configurations.

February 22, 2026

$The sequence is "xxx333xx". The third digit cannot be 3, so it must be 4 (1 choice). The seventh digit must be 4, and the last digit can be either 3 or 4, giving \(1 \times 2 = 2\) configurations.$

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If "333" starts at position 3:
The sequence is "xx333xxx". The second digit cannot be 3, so it must be 4 (1 choice). The sixth digit must be 4, and the last 2 digits can be either 3 or 4, giving \(1 \times 2^2 = 4\) configurations.
If "333" starts at position 4:
If "333" starts at position 5:
The sequence is "xxxx333x". The fourth digit must be 4 (1 choice), and the eighth digit can be either 3 or 4 (2 choices). The first 3 digits can be either 3 or 4, but must not form "333", which leaves \(2^3 - 1 = 7\) configurations
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