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Prove that $n \choose k$ $= \frac{P(n,k)}{k!}$

First of all, what does $P(n,k)$ represent? Then how do I got about this algebraically?

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    $P(n,k)$ is one of the common shorthands for the [Falling Factorial](https://en.wikipedia.org/wiki/Falling_factorial), $P(n,k)=\underbrace{n(n-1)(n-2)\cdots(n-k+1)}_{k~\text{terms}}$. In order to give any additional advice, we need to first know what definition of $\binom{n}{k}$ you were given as it has many different initial definitions which one eventually proves are all equivalent such as is the goal of this exercise. – 2017-01-19

3 Answers 3

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$P(n, k)$ represents the number of ways of removing, with order, $k$ objects from a bag of $n$ objects.

Having that said, the equality should hold almost immediately. You can check the spoiler below if you are really stuck:

$n\choose{k}$ is the number of groups of $k$ elements. Those can be obtained by removing $k$ objects from the bag of $n$ objects in $P(n, k)$ different ways. But removing first $a$ and then $b$ or first $b$ and then $a$ won't matter for the final group, thus you must identify all different orders of extraction that will create the same group. Since you are taking $k$ objects, they can be taken in $k!$ different ways.

Another way to look at it:

${n\choose{k}} = \frac{P(n, k)}{k!} \iff k!{n\choose{k}} = P(n, k)$. The LHS is the number of ways to create a bag of $k$ elements and then ordering them in $k!$ different ways. The RHS is directly the number of ways to create ordered extractions of $k$ objects from $n$.

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$P(n,k) = n!/(n-k)!$ is the number of "permutations of $n$ things taken $k$ at a time."

$C(n,k) = {n \choose k} = \frac{n!}{k! \times (n-k)!}$ is the number of "combinations of $n$ things taken $k$ at a time."

So the proof is obvious. Intuitively, an unordered combination can be 'arranged' in $k!$ ways to get an ordered permutation.

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I'm going to assume that $P(n,k)$ counts the number of $k$-permutations of a set of size $n$. This is also denoted $(n)_k$, or $n$ falling factorial $k$, and is equal to $\frac{n!}{(n-k)!}$. Algebraically, this is easy to prove.

You can also prove this combinatorially. Consider a $k$-permutation $(x_1,x_2...x_k)$ of $[n]=\{1,2...n\}$. Now consider the underlying set $\{x_1,x_2...x_k\}$. This is not a $1$ to $1$ relation- for example $(1,2,3)$ and $(3,1,2)$ both have the underlying set $\{1,2,3\}$. It is in fact a $k!$ to $1$ relation, as for each $k$-subset $S$ of $[n]$, there are $k!$ ways to permute it to get a $k$-permutation of $[n]$ with $S$ as its underlying set. You can factor the $k$-permutations into equivalence classes based on the relation of having the same underlying set, and each class has $k!$ elements. Thus, you divide the number of permutations by the size of the classes to get the number of classes, which is the number of $k$-subsets of $[n]$.