Hi Pam,

I am assuming that you can't repeat digits i the combination.

Since you are using the 7 digits 0, 1, 2, 3, 7, 8 and 9 there are 7 choices for the first digit in the combination and then 6 choices for the second digit and finally 5 choices for the third digit. Thus there are $7 \times 6 \times 5 = 210$ possible combinations. I can show you how to construct a list of the 210 possibilities.

Go to the Combinatorics Object Server at the University of Victoria in Victoria British Columbia and follow the link to Permutations and their restrictions. Select the radio button beside All k-permutations of {1,2,...n}. Set $n = 7$ and $k = 3$ and leave everything else as it is. Click on Generate! and you will be presented with a list of 210 triples. Unfortunately the triples use the digits 1 to 7 where you want 0, 1, 2, 3, 7, 8 and 9 and hence you will need to rewrite the list using

0 for 1
1 for 2
2 for 3
3 for 4
7 for 5
8 for 6 and
9 for 7.

I hope this helps,
Penny