[r-t] irregular leadheads
King, Peter R
peter.king at imperial.ac.uk
Thu Dec 2 15:05:59 UTC 2004
> 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
> to a valid (n-1)-cycle lead head. So there's the answer:
> (n-2)!.
>
That's what I thought and is true for n-1 prime but for Royal (for
example) there are only 6 regular lheads (and presumably each family of
irregular ones has only 6 members) to avoid the cycles of 3. Or am I
wrong?
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