Solving Combination Lock with 2nd and 3rd Numbers Differing by at Least 3

[fomenkoa]

Hey everyone

I have this for Discrete homework:
A lock has the numbers from 0 to 59 ...A combo is made up of 3 numbers...How many combos are possible if the 2nd and 3rd numbers have to differ by at least 3

The answer is whatever 60 times 58 times 57 is

I know why the 60 is there..because the 1st number can be any of the 60. How they got 58 times 57 I'm not exactly sure...

Can anyone help?
Anton

 

[vaishakh]

the first number can be anythng. for each first number there are 60 possiblities of second number. I think you understood this. For numbers 1 and 2 in the second lock, it is a must that 2 and 3, then 1, 3 and 4 should not be he number at third lock. similar is the case of numbers 60 and 59 appearing. for each other 56 cases of the second lock, there are 51 cases of third lock. Add the three different equation and then use algebra.

 

[fomenkoa]

the first number can be anythng. for each first number there are 60 possiblities of second number. I think you understood this. For numbers 1 and 2 in the second lock, it is a must that 2 and 3, then 1, 3 and 4 should not be he number at third lock. similar is the case of numbers 60 and 59 appearing. for each other 56 cases of the second lock, there are 51 cases of third lock. Add the three different equation and then use algebra.

I'm sorry...I don't quite understand this phrasing...could you repeat it please. Ok I know the first number in the combination can be any of the numbers 0-59. Then the second number in the combination can also be any of the 60 digits. However, the 3rd number needs to differ by at laest 3 from the 2nd number. How to get that in the form of the answer in the book, I don't know

The answer says to multiply 60 x 58 x 57 and taht gives u the amount of combinations. I am not sure why.

 

[vaishakh]

the first number can be anythng. for each first number there are 60 possiblities of second number. I think you understood this. For numbers 1 and 2 in the second lock, it is a must that 2 and 3, then 1, 3 and 4 should not be he number at third lock. similar is the case of numbers 60 and 59 appearing. for each other 56 cases of the second lock, there are 51 cases of third lock. Add the three different equation and then use algebra.

for the first number has no restriction on the second number. so for each number appearing on second lock there can be 60 cases of fist lock number.
but there are some complexity. if the number in the second lock is 1, or 60, then third lock should not feature 2 and 3, or 58 and 59 so that the difference is greater than 3. so for these two possibilities of second lock there can be 57 possibilties of third lock. (60*2*57) when the numbers are 2 or 59, the possibilities at third lock becomes 56. then for all other 56 numbers ecluding 1,2,59 and 60, the possibilities of third lock is 55(sorry, i wrote 51).
make three equations as the one i have given above.