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I would like to prove the formula for choosing $k$ elements out of $n$ and permuting them. I have been able to prove this by simple logic, as in there are n choices of items to place in the first position, $n-1$ choices for the second position, all the way to $(n - k + 1)$ choices for the last space.

If I wished to prove this by induction, would I need to manipulate both the variables, $n$ and $k$? Or is it possible to perform this proof by assuming $k$ is always some value less than or equal to n, or in the inductive case, equal or less than $n + 1$?

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    First suppose $n$ is fixed, and induct on $k$. Then induct on $n$, using/proving "Pascal's Formula" to do the induction. – 2017-02-25
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    $k$ cannot exceed $n$ (or $n+1$ in the inductive step); since the notion of picking more than $n$ elements out of $n$ would involve taking multisets and their permutations, which is a lot more complicated than simple permutations of distinct elements. – 2017-02-27
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    @Akay I have succesfully been able to show the steps when inducting on $k$. However, I run into a bit of trouble when attempting to induct on $n$. I have been trying to use pascals identity, or prove it, but I am not exactly sure what the inductive step would be. – 2017-03-01

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