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For the first number, there are 10 possible choices.
For the second number, there are 9 possible remaining choices.
For the third number, there are 8 possible remaining choices.
Therefore, for selecting three numbers,
. . there are: 10 \times 9 \times 8 \:=\:720 possible "combinations".
this is the email i got, but is 720 the answer for all of the "combinations", or just the ones begining with 1??
I AM CONFUSED

If you were to have the set, let's say S, containing all the integers 09, then there are 10 possible elements from which to choose. Now upon picking the second number there are 9 elements from which to choose. This assumes that there are no repeats. And finally as you said, there are 8 elements to from which to choose the third time picking. This shows that there are 720 distinct combinations or permutations of choosing three elements from this set.
They don't have to start with 1.
2,1,3 2,1,4... 2,1,9
2,3,1, 2,3,4... 2,3,9
etc.