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### Ben Evans says

This is an interesting permutation because 2 numbers repeat.

In this number, we can use 0 twice and 1 twice. We have the following number set (8, 3, 1,1,0,0)

So this looks like:

All combinations with both 8 or 3 with one of those numbers in the first position: (20)

83xx (8300, 8301, 8310, 8311)

38xx (3800, 3801, 3810, 3811)

8x3x (8030, 8031, 8130, 8131)

8xx3 (8003, 8013, 8103, 8113)

3x8x (3080, 3081, 3180, 3181)

1830, 1831 1380, 1381, 1803, 1813, 1308, 1318 (8 or 3 in the second position both numbers) (8)

, 1083, 1183,1038, 1138 (8 or 3 in the 3rd position both numbers) (4)

All numbers with a single 8 or 3 (12)

1800, 1801, 1810, 1300, 1301, 1310, 1080, 1081, 1030, 1031, 1180, 1130

Without zero in the front there are 44 combinations.

With zero in the first position there are a total of 68 combinations.

### Road Trip Amy says

That depends. Do you just mean the unique digits (0, 1, 3, 8) or do you mean all digits, with some repeated (0, 0, 1, 1, 3, 8)? If you mean the former, then I think the answer is:

4 x 3 x 2 x 1 = 24

There are 4 possible numbers, and 4 spots in the number you want to form. For the first digit you have 4 options, then for the second digit you have 3 remaining options, then for the third digit you have 2 remaining options, and only one remaining option for the final digit.