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The Science of Registration

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While organ registration is first and foremost an art, to be practised (hopefully) with musical sensitivity and good taste, there is an interesting mathematical side to it too. Now, if the mere mention of maths is enough to bring you out in a cold sweat, please don't panic just yet. There is no calculus or quadratic equations in what follows - just basic primary school arithmetic.

The relevant branch of mathematics is known as "permutations and combinations" or "perms and coms" as we used to call it when I was at school. To take permutations first, these are concerned with arranging things. If you haven't come across them before, a simple example will give you the idea. Suppose you have a small bookshelf which holds 6 books. How many different ways are there of arranging the books on the shelf?

The answer isn't hard to find if the problem is broken down into stages. Starting with an empty bookshelf, how many ways are there of choosing a book to go in the first position? Six, of course! What about the second position? One book is now used, so there are only five books left to choose from. Each of the six first choices gives a different five second choices, so the total number of ways of choosing books for both the first and second positions is
6 x 5 = 30

If we go one stage farther, the number of different ways of arranging books in the first three positions is

6 x 5 x 4 = 120

We can now see that the number of ways of arranging all six books is

6 x 5 x 4 x 3 x 2 x 1 = 720

That's a surprisingly large number of permutations, considering we are only dealing with six books, but if we increase the number of books, the number of permutations rises very dramatically:

7 books      5,040
8 books     40,320
9 books    362,880
10 books 3,628,800

So much for permutations. Combinations are the number of ways of selecting things. Again, a simple example will make the idea clear. Suppose, from our set of six books mentioned earlier, we wish to choose three to take on holiday. How many different ways are there of doing this? Following a similar line of reasoning we can say that there are six ways of choosing the first book, five of choosing the second, and four of choosing the third. This gives us

6 x 5 x 4 = 120.

However, this is not the whole story. With permutations,the order of the books is important, with combinations it isn't. Choosing books A,D and E to take on holiday is just the same as choosing books D,E and A, so we must adjust our answer (making it smaller)  to allow for this. To do this we have to think of the number of permutations of 3 books, which we now know is

3 x 2 x 1 = 6

To get the answer we divide our first number by our second number:

120 / 6 = 20

So there are 20 ways of choosing 3 items from 6 when the order isn't important, or 20 "combinations". Again if we were to increase the total number of books we would see the  number of combinations rising dramatically.

Strangely enough, the words permutation and combination are often used incorrectly in everyday speech. Take the case of the so-called combination lock. The order of the digits is important, so it should really be called a permutation lock, while the so-called permutations of the football pools are actually combinations. I'm pleased to say organists are a lot more logical in their thinking, and the combinations we refer to in connection with organ registration (yes, I'm getting to the point at last!) are true combinations in the mathematical sense.

Using the simple tools above, we now have a means of working out the total number of possible combinations for any organ. If, for example, we have a small  instrument with 6 speaking stops, we already know (from the holiday books example above) that there are twenty possible combinations involving 3 stops. To get the total number of combinations, using any number of stops, we need to repeat the calculation for 1,2,4,5 and 6 stops drawn, and then add all the answers together:

One stop combinations:  (6)/(1) = 6
Two stop combinations:  (6 X 5)/(2 x 1) = 15
Three stop combinations: (6 x 5 x 4)/(3 x 2 x 1) = 20
Four stop combinations:  (6 x 5 x 4 x 3)/(4 x 3 x 2 x 1) = 15
Five stop combinations:  (6 x 5 x 4 x 3 x 2)/(5 x 4 x 3 x 2 x 1) = 6
Six stop combinations:  (6 x 5 x 4 x 3 x 2 x 1)/(6 x 5 x 4 x 3 x 2 x 1) = 1
TOTAL    63

As you can see, the numbers fall into a symmetrical pattern, and this agrees with common sense, because choosing four stops to draw is equivalent to choosing two stops not to draw.

If we are interested only in the total number of combinations, there is a quicker way to get at the answer. We consider each stop control in turn and remember it can be either ON or OFF. If we just look at one, there are two possible states. If we look at two together, the number of possible states is

2 x 2 = 4

Finally, if we consider all six stops, the total number is

2 x 2 x 2 x 2 x 2 x 2 = 64.

At first sight, our two answers don't quite agree, but this is because the second method of calculation includes the case of zero stops drawn. If we agree not to include this as a combination, the answer is 63 in both cases. Yes, an organ with only 6 speaking stops gives you a total of 63 possible combinations! (Not all may be musically pleasing, of course). But if this number surprises you, see what happens when the number of speaking stops increases:

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 (approximately)

(Don't try this sum for the Wanamaker organ or your calculator may explode!)

It's mind boggling stuff, and one of the reasons playing the pipe organ is so fascinating. Even if your instrument isn't a particularly large one, and even if you've been playing it for years, there are sure to be plenty of combinations which you haven't even heard yet! Perhaps you are inhibited from experimenting due to preconceived notions of what is "allowed"? Why not let an inquisitive child loose on your instrument? Let them pull stops at random and see what they come up with.


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    This is the clearest explanation on permutations and combinations I have encountered (and my masters project involved combinatorics) - so intuitive! Thank you!

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