hmmm... 8 choices for High/Low
8 choices for vtrac strings (using V=vtrac, a=first number V represents, b=second number, for example V1=a0 or b5)
VaVaVa, VaVaVb, VaVbVa, VaVbVb, VbVaVa, VbVaVb, VbVbVa, VbVbVb
8 choices for odd/even
guess a good rule of thumb would be that any filter that has a binary component (2 choices, H/L, VaVb, O/E) will have 8 resultant possibilities in a pick 3 scenario and 16 in a pick 4... (HHHH,HHHL,HHLH,HHLL,HLHH,HLHL,HLLH,HLLL,LHHH,LHHL,LHLH,LHLL,LLHH,LLHL,LLLH,L)LLL
Am I getting this so far?