Fundamental Counting Principle problem

[PiRsq]

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?

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?

 

[HallsofIvy]

Not quite. If the first number is 57, 58 or 59, then the next number CAN'T be "3 larger" but only 3 less. If the first number is 0, 1, or 2, the next number CAN'T be "3 less but only 3 more.
That is, if the first number is 3 to 56 (54 numbers) then there are 2 possible second number but if the first number is 0, 1, 2, 57, 58, or 59, then there is only 1 possible second number. There are 54*2+ 3+ 3= 114 possible two digit combinations for the first two numbers. There are 114*60= 6840 such three digit combinations.

 

[chroot]

Halls,

While he didn't explicity state it, I would think that you can go "over the top" and consider 59 and 2 to differ by 3.

Pi,

I think you're almost correct. However, if you choose one of 60 numbers, then choose one of the 59 remaining, there are 58 left -- NOT 57.

- Warren

 

[PiRsq]

Thanks for the replies guys. Hall I don't understand why you did 54*2 and added 6