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
Jul 8th 2006, 09:22 AM
If you were to have the set, let's say S, containing all the integers 0-9, 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.