[r-t] Bobs only Stedman Triples

[Philip Saddleton]

Stephen Beckingham said  on 26/06/2012 08:21:
>> From: bex280 at hotmail.com
>>> The peal of Erin requires 840 sixes.
>>>
>>> That makes 280 Q-sets. Now have to decide which ones are plained and with
>>> ones are bobbed.
>>
>> If it were that straightforward I would have thought someone's computer would have already done it.
>>
>> SJB
>
> Sorry, clearly I meant:
> 1234567 is bobbed, and 1234675 is plained. This means that 1234756 can not be had AT A SIX END, but surely 2314756 or 3124756 can be, thus making the q set rule redundant in this case.

However the number of bobbed sixes ending in 567 must be the same as the 
number of bobbed sixes ending in 675, or we have the wrong number of 6x75s.

That makes 70 sets of 12 sixes, for each of which we can choose
1 way each of bobbing all sixes or plaining all sixes
3 x 3 x 3 ways each of bobbing one or two with each of the three bells 
behind (depending on the bell in 4th's place)

So only 56 ^ 70 possibilities to try.