Richard Smith richard at ex-parrot.com
Thu Dec 2 14:14:06 UTC 2004

```Peter King asks:

> How many leadheads are there on n bells which (keep the treble fixed)

There are n! rows on n bells and, in total, (n-1)! lead
heads which form the permutation group S_(n-1).  So we now
need to know how many elements of S_(n-1) are n-1 cycles.
As they are n-1 cycles, we can write down the place bell
order for each of them starting from some arbitrary choice
of bells (say the tenor).  We then find that there are
(n-2)! possible place bell orders each of which corresponds
(n-2)!.

> Regular leadheads are labelled a, b, c etc., is there a similar

The are names in common usage for Minor (and, I believe,
occur in symmetric methods with 2nds or 6ths place lead
ends.

In Minor, they form four sequences

G H J K  (2nds place regular methods)
L M N O  (6ths place regular methods)
P Q R S  (2nds place irregular methods)
T U V W  (6ths place irregular methods)

The regular symbols correspond to the more modern lower case
letters as follows:

G=a, H=b, J=e, K=f, L=g, M=h, N=l, O=m

Why are there so few symbols?  Wouldn't we expect there to
be 2 x (n-2)! = 48?  (That is, 24 lead heads reached either
with 2nds or 6ths.)

The reduction is becaues, if we restrict ourselves to
symmetric methods, the lead end row must be conjugate to
this must have (n-2)/2 pairs swapping (at an even stage).
This gives the number of lead ends as

(n-1)!
-------------------  =  (n-1)(n-3)...3.
(n-2)/2   n-2
2           --- !
2

Or 15 for Minor.  Five of these are the regular lead ends
and produce the regular lead end codes.  Another seven (not
eight as might be expected -- S and V share a common lead
end) produce the irregular lead end codes.  The remaining
three cannot produce a 5-cycle using either 2nds or 6ths