An alphabetic lock uses 3 distinct letters chosen from 4 available letters. How many possible codes?

An alphabetic lock uses 3 distinct letters chosen from 4 available letters. How many possible codes?

• A. 24
• B. 12
• C. 6
• D. 8

Answer: (A)

Since the code uses 3 distinct letters from a set of 4 and the order matters, you’re counting permutations of 3 items from 4. For the first position you have 4 choices; for the second, 3 remaining choices; for the third, 2 remaining choices. Multiply: 4 × 3 × 2 = 24. This is 4P3 = 4!/(4−3)! = 24. So there are 24 possible codes.

Since the code uses 3 distinct letters from a set of 4 and the order matters, you’re counting permutations of 3 items from 4. For the first position you have 4 choices; for the second, 3 remaining choices; for the third, 2 remaining choices. Multiply: 4 × 3 × 2 = 24. This is 4P3 = 4!/(4−3)! = 24. So there are 24 possible codes.