SOLUTION: A pin code is a sequence of three numbers, each of which is between 1 and 9. If none of the numbers in the sequence can repeat, how many combinations for the code are possible?

I'm not sure what you mean by "between" -- inclusive or exclusive? 
1 and 9 are not "between" 1 and 9, Do you mean including 1 and 9 
or just the seven numbers 2,3,4,5,6,7, and 8?

If you are including 1 and 9, then,

Choose the first number any of 9 ways.
For ever one of those 9 ways to choose the first number, 
there are 8 ways left to choose the second number,
So that 9*8 or 72 ways to choose the first two numbers.
For each of those 72 ways to choose the first two numbers,
there are 7 ways left to choose the third number, so that's
9*8*7 or 72*7 or 504 ways.

If you are NOT including 1 and 9, but only the 7 numbers in
between 1 and 9, that is, 2,3,4,5,6,7,and 8 then,

Choose the first number any of 7 ways.
For ever one of those 7 ways to choose the first number, 
there are 6 ways left to choose the second number,
So that 7*6 or 42 ways to choose the first two numbers.
For each of those 42 ways to choose the first two numbers,
there are 5 ways left to choose the third number, so that's
7*6*5 or 42*5 or 210 ways.

So, the answer is either 504 or 210 depending on what you mean
by "between 1 and 9"
 
Edwin