Combinatorial Correctness of one-to-one functions Let $\lbrack k\rbrack$ be the set of integers $\{1, 2, \ldots, k\}$. What is the number of one-to-one functions from $m$ to $n$ if $m \leq n$? My answer is: $\dfrac{n!}{(n-m)!}$ 
My reasoning is the following:
We have an $m$-step, independent process:
Step 1: choose the first $m \in \lbrack m \rbrack$ to be mapped to a $n \in \lbrack n \rbrack$
There are $n$ choices.
Step 2: choose the second $m \in \lbrack m \rbrack$ to be mapped to a $n \in \lbrack n \rbrack$ There are $n-1$ choices here since we cannot map to the $n$ in the previous step (as we must count one-to-one functions) 
Repeat this until for $1, 2, \ldots m$. This is $n(n-1)(n-2) \ldots (n-m+1) = \dfrac{n!}{(n-m)!}$


*

*Is this correct?

*Is my reasoning correct?


It is OK, modulo minor problems. You don't have to select the $m$'s, just go with the natural order 1, 2, ... Take a look at the notation suggested by Knuth et al in "Concrete Mathematics", it really does clean up much clutter.
