f(1) can be any element of the codomain {1,2,3,4,5,6,7,8,9,10}: 10 choices. Next, f(2) is not equal to f(1), so f(2) can be any element of {1,2,3,4,5,6,7,8,9,10} except for f(1): 9 choices. Similarly, there are 8 choices for f(3), ..., and 3 choices for f(8).How many injective maps are there?

f : {1,2,3,4,5,6,7,8} → {1,2,3,4,5,6,7,8,9,10}