A lock's combination consists of three digits. The first digit is a power of 2, the second digit is a prime number, and the last digit is a factor of 10. How many unique combinations fit this criteria?
The first digit is a power of 2, so $2^n$, which the possibilities is $n=\{0,1,2,3\}$ since a lock has numbers 0-9.
The second digit is a prime number so $\{2,3,5,7\}$
And the last digit is a factor of 10 so $\{1,2,5\}$.
Figured out my mistake.
Best Answer
There are four options for the first digit, four options for the second digit, and three options for the third digit. So there are $4 \cdot 4 \cdot 3 = 48$ possible combinations.