Combination lock probability

Ok, lets try this.
Write the numbers 0 to 59 in a line.

Next line put an X under the first number in the combination, 0.
The next two numbers 1 & 2 are NOT three away so put X's under those also.

Next line put an X under the next first number in the combination, 1.
0,2 &3 are NOT three away so put X's under those also.

Next line put an X under the next first number in the combination, 2.
0, 1, 3 & 4 are NOT three away so put X's under those also.

Next line put an X under the next first number in the combination, 3.
1, 2, 4, & 5 are NOT three away so put X's under those also.

Next line put an X under the next first number in the combination, 4.
2, 3, 5 & 6 are NOT three away so put X's under those also.

Continue doing that till the last four lines are

Next line put an X under the next first number in the combination, 56.
54, 55, 57, & 58 are NOT three away so put X's under those also.

Next line put an X under the next first number in the combination, 57.
55, 56, 58 & 59 are NOT three away so put X's under those also.

Next line put an X under the next first number in the combination, 58.
56, 57, & 59 are NOT three away so put X's under those also.

Next line put an X under the next first number in the combination, 59.
57 & 58 are NOT three away so put X's under those also.

Now you have accounted for every first number in the combination.

The squares without X's are under the possible second numbers in the combinations. Count them and multiply by 60 for the third number.
------------------
Gene
 
You have 60 lines.
Each line has 60 squares.
No line has more than 5 X's
That leaves at least 55 squares in each line without an X.
 
You have 60 lines. Yes
Each line has 60 squares. Yes
No line has more than 5 X's ?
That leaves at least 55 squares in each line without an X. ^?
 
The top left corner looks like
Code: 
0 1 2 3 4 5 6 7 8 9...
X X X
X X X X
X X X X X
  X X X X X
    X X X X X
      X X X X X
        X X X X X
          X X X X X 
.
.
.
 
I had the silly thought that you might actually read the part where the quote was. The entire quote is
Now you have accounted for every first number in the combination.
The squares without X's are under the possible second numbers in the combinations. Count them and multiply by 60 for the third number.
Click to expand...

What that means is that what you counted combined the first number and the second number.
 
So the number for the second number is 299
I divide by 60 of course
then it is: 59
 
You can't do it that way. Depending on which number you use as the first there are different possibilities for the second. Counting the table adds up all the possibilities. It is
57+56+(56*55)+56+57 = 3306
for the first two numbers. 60 for the third number gives
3306 * 60 = 198360 for all three numbers.
 
the number of possibilities for the second number of the combination
it is 55 or 56 or 57 I think
 
That is like asking "What day is tomorrow?" If today is Friday tomorrow is Saturday. If today is Monday tomorrow is Tuesday. The answer changes.
If the first is 0 or 59 the second is indeed 57.
If the first is 1 or 58 the second is indeed 56
If the first any thing else the second is 55
Look at the %$@!^&# table.
The question was how many possible combinations are there? The answer is 198360. That has been answered several times in several ways.
I HAVE NO MORE TO SAY!!!!
 
Pingu, you've really ticked off Gene. I've never seen him like this before. This has gone on for 3 pages. Can we just end it?
 
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