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 Question from Marvin: 4 digit combinations 0 through 13 non-repeating I am designing a game, and I need away of figuring this out without having to write numerically all these numbers

Hi Marvin,

I assume order is important, for example 6, 4, 12, 7 is not the same as 12, 7, 4, 6. If so then you have 14 choices for the first integer to choose, any integer from 0 to 13. Once you have chosen the first integer you have 13 choices for the second integer. Thus there are $14/ times 13$ choices for the first two integers.

Whatever integers you have chosen for the first two there are 12 choices for the third integer and hence there are $14 \times 13 \times 12$ possibilities for the first three integers. Similarly there are $14 \times 13 \times 12 \times 11 = 24,024$ possibilities for a four integer combinations.

This is too many to list.

Penny