A plain course of Bob Doubles has forty rows, all different. Are there more rows we could ring? How many?

Asking that question is the same as asking ‘In how many different ways can you arrange five letters?’ Let’s find out.

One letter can be arranged in just one way:

 A

If we add a second letter, we can either put it in front of the first one, or behind it:

 AB    

or

 BA    

We had 1 arrangement for 1 letter, and we can add the second letter in one of two positions.

That gives us 1x2 arrangements, which is, of course, equal to 2 arrangements.

Now, let’s add a third letter.

We have two arrangements of our first two letters. In each of those arrangements, we can add the third letter in one of three positions:

• in front

• in the middle

•  at the back

Let’s do each of those with AB:     

            CAB 

            ACB

            ABC

and now, with BA:     

            CBA

            BCA

            BAC

We had 1x2 arrangements of two letters. In each of those arrangements we can add the third letter in one of three positions. That gives us 1x2x3 arrangements, which is, of course the 6 arrangements of three letters that you can see.

Lets add a fourth letter, but lets save ourselves some time and do some maths, Frankie!

We had 1x2x3 = 6 arrangements of three letters. In each of those six arrangements, we can put the new letter in one of 4 positions (either 1st, 2nd, 3rd or last)

That means we must have 1x2x3x4 = 24 arrangements of 4 letters. You can draw them all out if you feel like it, or you could just write out a plain course of Plain Bob Minimus (4 bells, it’s Plain Hunt on 4, with 2nds made at the end of each lead; try it!). It has 3 leads of 8 rows, 24 rows in total, and I challenge you either to find a row that isn’t there, or find one that repeats.

You should now be able to see the pattern, and so, I’m going to claim that on five bells, there are 1x2x3x4x5 = 120 possible rows.

So far, we’ve only managed to ring 40 of them. Where do we get the remaining 80 from? Go Grandsire…