Question 309212
<font face="Garamond" size="+2">


There are a total of 10 digits counting zero.  So if you must exclude zero for the first digit, then there are 9 possibilities for the first digit.


For each of those 9, and since I can now use zero but cannot repeat the first digit, there are 9 choices (the 9 I had to begin with plus one for the zero becoming available, and minus one for the one I used on the first digit).  9 times 9 = 81


For each of those 81 combinations, I now have 8 choices -- the nine I had when I chose the second digit, minus the one that I chose.  81 times 8


For the 4th digit, I have 7 choices.  81 times 8 times 7


And then for the 5th and 6th digits I have 6 and 5 choices:


81 times 8 times 7 times 6 times 5 = (you do your own arithmetic)


John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>