Permutation is a big word, and sounds complicated, but it really isn't. A permutation is an arrangement of objects. For example, suppose I have a square, a circle, and a triangle. If I put arrange them in the order I just listed, that is a permutation. If I arranged them with the triangle first and the square last, that would be another permutation. If we wanted to, we could list all the permutations. To simplify it, we can call the square "S," the circle "C," and the triangle "T." Then the arrangements are:

SCT, STC, CST, CTS, TSC, TCS.

There are six arrangements. How did I make sure that I got them all? First, I listed all the arrangements that had the square first, then the ones that had the circle first, then the arrangements that had the triangle first. Suppose I added a pentagon (P) into the mix. What would that do? Well, we could take that P and insert it into any position of any of the six permutations shown above. That would result in a total of 24 permutations.

Here's another example. Suppose we wanted to make arrangements of our square, circle, triangle, and pentagon but we were only arranging two out of the four. How many arrangements would there be?

SC, ST, SP, CS, CT, CP, TS, TC, TP, PS, PC, PT, for a total of 12 permuations.

One of the things we're often interested in is not what the permutations are, but how many there are. And for that, it would be nice to have a formula. And there is one!

If n is the number of objects, and r is the number of objects you are selecting arranging, here's your formula:

_nP_r =

This is read as follows: "Permutations of n things r at a time is n factorial divided by n minus r factorial."

Do you remember what a factorial is? A factorial is what you get when you multiply an integer by all the integers between it and 1. For example, 5! = 5 x 4 x 3 x 2 x 1 = 120.

So if we have four objects (square, circle, triangle, and pentagon), and we're choosing two of them,

₄P₂ = = = 12, which is exactly what we calculated.

Let's try another example. Suppose you want to take the letters "MADELYN" and pick three of them to arrange, how many permutations are there?

We could list all the permutations, but in this case that would be a lot more writing. So let's use the formula. The number of objects (letters) is 7, and we're picking three of them, so we want ₇P₃.

₇P₃ = = = 210. Wow! I'm glad we didn't try to count those by hand!

I should point out that MADELYN doesn't have any repeated letters; if it did, the problem would be a bit different. Why? Suppose the word was "MADALYN". Now you could have "MAA" and "MAA" as two arrangements (in the first arrangement, we use the first A first, and in the second arrangement, we use it last). And yet, they are indistinguishable - without me telling you which A went where, you'd never know, would you? So we wouldn't want to count those twice.

We'll explore how to handle that situation in the next section.

1.

What does the word "permutation" mean?

2.

If you have 5 books on a shelf, how many ways are there to arrange them?

3.

If you have 5 books on a shelf, and you want to put two of them on another shelf, in how many ways can you do this?

4.

In how many ways can you arrange n books?

5.

Calculate ₁₀P₅

6.

In how many ways can you arrange the letters of the word MATH?

7.

In how many ways can you arrange three letters from the word MATH?

8.

Calculate ₅P₅