The dial on a 3 number combination lock contains markings to represent the numbers from 0 to 59. How many combinations are possible if the first and second numbers differ by 3?(adsbygoogle = window.adsbygoogle || []).push({});

What I did was:

1st number: It can be any of the 60 numbers (if we take 0 also as a #)

2nd number: I think since there are two possibilities, either 3 greater than 1st # or 3 less

3rd number: Since you've already take a number for the first one, and you must choose either of 3 less or 3 greater than the first number as the 2nd #, you must in the end have 57 #'s left to choose from

Therefore the answer I think is 60 x 2 x 57

Is that right?

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# Fundamental Counting Principle problem

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