[r-t] irregular leadheads

[Richard Smith]

Andrew Tyler wrote:

> 1. There are eight groups of seven lead heads that yield
> 2nds place methods.

Eight groups of six lead heads, surely?  One of them is
rounds and so gives a 1-lead plain course.

> 2. There are eight groups of seven lead heads that yield
> 8ths place methods.

And similarly, eight groups of seven leads (six if you
exclude rounds) that yield 4ths placce methods, and another
eight groups of seven yielding 6ths place methods. Adding
them all up gives 8x6x4 = 192 which is consistent with
2^(n/2-1) (n/2)!.

If we want to ignore 4ths and 6ths place lead heads, the
answer becomes 2^(n/2) (n/2-1)! (or 96 on eight bells).

Richard