So, did you come up with a working rule to describe the pattern we looked at last time? Here’s what I came up with:

As we saw last time, the “root” of the tree diagram (the first column) shows \(\mathtt{_{4}P_1}\), which is the number of permutations of 1 card chosen from 4. The first and second columns combined show \(\mathtt{_{4}P_2}\), the number of permutations of 2 cards chosen from 4. So, to determine \(\mathtt{_{n}P_r}\), according to this pattern, we start with \(\mathtt{n}\) and then multiply \(\mathtt{(n-1)(n-2)}\) and so on until we reach \(\mathtt{n-(r-1)}\).

The number of permutations of, say, 3 items chosen from 5, then, would be \[\mathtt{_{5}P_3=5\cdot (5-1)(5-2)=60}\]

This is a nice rule that works every time for permutations of \(\mathtt{r}\) things chosen from \(\mathtt{n}\) things. It can even be represented a little more ‘mathily’ as \[\mathtt{_{n}P_r=\prod_{k=0}^{r-1}(n-k)}\]

So let’s move on to the “legal” formula for \(\mathtt{_{n}P_r}\). A quick sidebar on notation, though, which we’ll need in a moment.

When we count the number of permutations at the end of a tree diagram, what we get is actually \(\mathtt{_{n}P_n}\). In our example, that’s \(\mathtt{_{4}P_4}\). The way we write this amount is with an exclamation mark: \(\mathtt{n!}\), or, in our case, \(\mathtt{4!}\) What \(\mathtt{4!}\) means is \(\mathtt{4\times(4-1)\times(4-2)\times(4-3)}\) according to our rule above, or just \(\mathtt{4\times3\times2\times1}\). And \(\mathtt{3!}\) is \(\mathtt{3\times(3-1)\times(3-2)}\), or just \(\mathtt{3\times2\times1}\).

In general, we can say that \(\mathtt{n!=n\times(n-1)!}\) So, for example, \(\mathtt{4!=4\times3!}\) etc. And since this means that \(\mathtt{1!=1\times(1-1)!}\), that means that \(\mathtt{0!=1}\).

So, for the tree diagram, \(\mathtt{_{4}P_4}\) means multiplying all the way to the right by \(\mathtt{n!}\). But if we’re interested in the number of arrangements of \(\mathtt{r}\) cards chosen from \(\mathtt{n}\) cards, then we need to come back to the left by \(\mathtt{(n-r)!}\) And since moving right is multiplying, moving left is dividing.

The division we need is not immediately obvious, but if you study the tree diagram above, I think it’ll make sense. This gives us, finally, the “legal” formula for the number of permutations of \(\mathtt{r}\) items from \(\mathtt{n}\) items: \[\mathtt{_{n}P_r=\frac{n!}{(n-r)!}}\]