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.


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.

  • $\begingroup$ Wait so that was all? no need for the combination or permutation formulas? $\endgroup$ – Killercamin Feb 14 '17 at 21:07
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    $\begingroup$ @FernandoFernandez no need for those. It's just a simple application of the Multiplication Principle. $\endgroup$ – Harry Feb 14 '17 at 21:10
  • $\begingroup$ Thanks for the explanation and quick response $\endgroup$ – Killercamin Feb 14 '17 at 21:10

You're not rearranging anything, so you don't need permutations.

I'll pose a similar question: How many two-digit numbers have the first digit odd, and the second digit even?

There are five odd digits, and five even digits, so the total number of two-digit numbers that meet the criteria is $5 \times 5 = 25$.

Can you take it from here?

  • $\begingroup$ Yeah I just figured it out thanks $\endgroup$ – Killercamin Feb 14 '17 at 21:10

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