A slight digression, with some very simple (I hope) maths, to consider how many variations of the registrations can be obtained with just a few stops.

Suppose I have a very small organ with just 6 speaking stops on it.

Let us list all the possible combinations for drawing 1, 2, 3, 4, 5, or 6 stops, then add them up.

We obviously have 6 alternatives for our choice of the first stop, so that leaves 5 for the second choice. So there are 6x5 ways of choosing 2 stops from our 6, which equates to 30.

But this is technically a permutation and we want a combination. What’s the difference?

If we choose Stop A and Stop B, we will also have another selection of Stop B and Stop A, in which case we have 2 sets the same.

So the number of ways of choosing 2 different stops from a set of 6 is:-

(6x5) divided by (2X1), which equals

**15**.

Does that surprise you?

Let us list all the possible combinations for drawing 1, 2, 3, 4, 5, or 6 stops, then add them up.

One stop combinations - 6 x1 = 6

Two stop combinations - (6x5) / (2x1) = 15

Three stop combinations - (6x5 x4) / (3x2x1) = 20

Four stop combinations - (6x5x4x3) / (4x3x2x1) = 15

Five stop combinations - (6x5x4x3x2) / (5x4x3x2x1) = 6

Six stop combinations - (6x5x4x3x2x1) / (6x5x4x3x2x1) = 1

**Total = 63**As you can see the individual totals fall into a symmetrical pattern, and that is logical, because

**choosing 4 stops is equivalent to choosing 2 stops not to pull out**.

If we are interested only in the total number of combinations there is a quicker way to get the answer.

If we consider each stop in turn, they are either

**on or off**.

If we look at one stop there are only 2 possible results (on or off).

If we look at 2 stops together, the number of possible combinations is 2 x 2 = 4.

Now if we consider all 6 stops, the number of possible combinations is:

2 x 2 x 2 x 2 x 2 x 2 =

**64**.

Ah, I hear you say, “That is not quite correct”.

The reason is that the second method also includes the case where zero stops are drawn out!

So by omitting that meaningless, in our situation, alternative both results give us 63.

Yes, that’s right an organ with only 6 speaking stops can give you a choice of 63 different registrations. I don’t know that everyone of them would be musically pleasing though!

If that surprises you see what happens with more stops on the organ.

10 stops 1,023 combinations

20 stops 1,048,575 combinations

30 stops 1,073,741,823 combinations

40 stops 1,099,511,627,780 combinations

The formula for working these figures out is

**2**^{n} - 1, where

**n** is the total number of speaking stops available.

You can see how this is derived by clicking this link:

http://www.ar-group.org/smforum/index.php?topic=3025.0This is mind blowing, but it shows us one of the reasons that playing the organ is so fascinating.

To take a recent case that I posted last week, the organ in Odiham Parish Church has 22 stops. This means that that particular organ has well over 1 million variations to possible registrations. So how long would it take to work through them all?

You can view that posting by clicking this link:

http://www.ar-group.org/smforum/index.php?topic=3017.0It means that even if you have

**quite a small organ**, and our

**Yamaha AR organs are larger than most**, in this respect, and even if you have been playing it for years, there are sure to be many registration combinations that you have not yet discovered.

Do our inhibitions or preconceptions of what is allowed prevent us from exploring?

Why not let your inquisitive nature loose and

**pull out a few of those other stops!**In the next Reply we see why the organ has the largest range of any instrument.

Peter