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With 3 digit numbers there are 10 digits (0, 1, ..., 9) from which to select 3 different digits. This can be done in:

10C3 = 10! / (10-3)!3! = 120 ways.

For a set with n items, r of them can be selected (permuted) in:

n x (n-1) x ... x (n - (r +1)) = n! / (n-r)! = nPr

ways. However, if the order of selection does not matter (just the combination of the r items) then the selected items can be chosen in

r x (r-1) x ... x 1 = r!

ways. Meaning there are

nPr / r! = nCr = n! / (n-r)!r!

combinations possible.

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Q: How many different 3 digit combinations can there be if the digits are distinct?
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