For permutations, is there any mathematical representation that $(n - r)!$ has other than being used to take eliminate the tail of the numerator $n!$ so that only the number of "choices" we make are left in the end?
I saw someone post online about how $(n - r)!$ represent the number of ways we can arrange the numbers we don't care about, is that true?
Thanks!
Yes, it is true. Let us take $n=7,r=4$ for example. There are $7!$ ways to arrange all $7$ items in a line. If I only want three of them, there are $\frac {7!}{(7-4)!}=7\cdot 6 \cdot 5$ ways to arrange them in a line. One way to see this is to say there are $7$ choices for the first, $6$ choices for the second, and $5$ choices for the third. Another way is to say there are $7!$ ways to put them in a line, but I don't care how the last four are arranged because I will ignore them. There are $4!$ ways to arrange the last $4$, so there are $\frac {7!}{4!}$ ways to arrange the first three.
Consider the following way to find the number of arrangements. Suppose you have $r$ distinguishable apples and $n-r$ indistinguishable oranges. Let's say $x$ is the number of distinct ways to arrange them. Now for every such arrangement, there are $(n-r)!$ ways to label the oranges from $1$ to $n-r$. So the number of distinguishable arrangements after labelling the oranges is exactly $x\cdot(n-r)!$, but it is also $n!$, since now you're just arranging $n$ distinguishable fruits. So you get $x = n!/(n-r)!$.
