FUNCTION in easyer format

[ytayta555]

Have a good time !

I have this function :
=AND(COUNT(Z4;Z5;Z6;Z7;Z8;Z9;Z10;Z11;Z12;Z13;Z14;Z15;Z16;Z17;Z18)<=1;COUNT(AA4;AA5;AA6;AA7;AA8;AA9;AA10;AA11;AA12;AA13;AA14;AA15;AA16;AA17;AA18)<=1;COUNT(AB4;AB5;AB6;AB7;AB8;AB9;AB10;AB11;AB12;AB13;AB14;AB15;AB16;AB17;AB18)<=1;COUNT(AC4;AC5;AC6;AC7;AC8;AC9;AC10;AC11;AC12;AC13;AC14;AC15;AC16;AC17;AC18)<=1;COUNT(AD4;AD5;AD6;AD7;AD8;AD9;AD10;AD11;AD12;AD13;AD14;AD15;AD16;AD17;AD18)<=1;COUNT(AE4;AE5;AE6;AE7;AE8;AE9;AE10;AE11;AE12;AE13;AE14;AE15;AE16;AE17;AE18)<=1;COUNT(AF4;AF5;AF6;AF7;AF8;AF9;AF10;AF11;AF12;AF13;AF14;AF15;AF16;AF17;AF18)<=1;COUNT(AG4;AG5;AG6;AG7;AG8;AG9;AG10;AG11;AG12;AG13;AG14;AG15;AG16;AG17;AG18)<=1;COUNT(AH4;AH5;AH6;AH7;AH8;AH9;AH10;AH11;AH12;AH13;AH14;AH15;AH16;AH17;AH18)<=1;COUNT(AI4;AI5;AI6;AI7;AI8;AI9;AI10;AI11;AI12;AI13;AI14;AI15;AI16;AI17;AI18)<=1;COUNT(AJ4;AJ5;AJ6;AJ7;AJ8;AJ9;AJ10;AJ11;AJ12;AJ13;AJ14;AJ15;AJ16;AJ17;AJ18)<=1;COUNT(AK4;AK5;AK6;AK7;AK8;AK9;AK10;AK11;AK12;AK13;AK14;AK15;AK16;AK17;AK18)<=1)

Can somebody give me an equivalent for this function , to be more
simply ?

1.) Every count function must work like an normal Count function, not
like COUNTIF function with criteria<=1 !
2.)Like this doesn't help me
(COUNT(AG4:AG18)<=1;COUNT(AK4:AK18)<=1;............), because this
function is for exemple , I must made and built milions af this
function , with references in combinatoric order !
3.) Every count function is a combination of 15 numbers , used for an
lotto statistic database , it doesn,t work to count
{ COUNT(AH4;AI4;AJ4;AK4)................., the drow is in an one
column , not in one row .

Thanks very much for your amability and for your time !

 

[Joel]

the code below is 234 characters which will work in excel. I replaced the
semicolon with commas.

=AND(COUNT(Z4:Z18)<=1,COUNT(AA4:AA18)<=1,COUNT(AB4:AB18)<=1,COUNT(AC4:AC18)<=1,COUNT(AD4;AD5:AD18)<=1,COUNT(AE4:AE18)<=1,COUNT(AF4:AF18)<=1,COUNT(AG4:AG18)<=1,COUNT(AH4:AH18)<=1,COUNT(AI4:AI18)<=1,COUNT(AJ4:AJ18)<=1,COUNT(AK4:AK18)<=1)

 

[ytayta555]

Thanks very much sir Joel , but , how I said :
2.)Like this doesn't help me
(COUNT(AG4:AG18)<=1;COUNT(AK4:AK18)<=1;............), because this
function is for exemple , I must made and built milions af this
function , with references in combinatoric order ! .............

I shall have 1;3;6 1;5;9 for eg .........for rows ; and I have
to built milions of
this kind of functions , with references in COMBINATORIC order ;nobody
could
to help me , I get some ideeas , but don't work :

=COUNT(INDEX(MATCH(ROW(B4:AH83),
{4,7,9,11,14,16,18,22,25,35,46,57,68,72,83},0)/
ISNUMBER(B4:AH83),0))

=SUMPRODUCT(ISNUMBER(MATCH(ROW(B4:AH83),
{4,7,9,11,14,16,18,22,25,35,46,57,68,72,83},0))*
ISNUMBER(B4:AH83))

=SUMPRODUCT(ISNUMBER(MATCH(ROW(B1:F5), {1,3,5},0))*(B1:F5<=1))
for this function posted of me in another thread :
=AND(COUNT(B1;B3;B5)<=1;COUNT(C1;C3;C3)<=1;COUNT(D1;D3;D5)<=1;COUNT(E1;E3;E­
5)<=1;COUNT(F1;F3;F5)<=1)

Many thanks

 

[Joel]

I think you need to do what you want using VBA instead of worksheet
functions. I can't tell exactly what you want from you vague examples. You
are looking for combinatorial function which can be coded in VBA. I've done
it before many times.

Need more info to help write this code.

 

[ytayta555]

Need more info to help write this code.

Thanks very much for reply , sir Joel

I'ts a long story . I have a lotto statistic database of 231 workbooks
now .In every wbook I have 65446 Count functions , with the 15
references
in combinatoric order (placed in column B ,row 91 to 65536 );
actually , I
use autofill method for every wbook to find which combination of
numbers
hasn't >=2 numbers in last continuous 33 draws (combined with an
Countif
function placed in Column BD ); an VBA macro loop trough Column BD and
if
the result of Countif function is >=33 , then copy antyre row and
paste it
in another wbook ;in every wbook in range A1:BB90 are
last 53 draws of this lottery ;this is the kind and wey of the querry
in this
database for to find the combinations of 15 numbers (from 90) which
has
in 33 continuous draws results of <2 numbers ;the autofill is from
B91:B65536
to BB91:BB65536 , and it take 2 minute time for every workbook to end
the querry
in every wbook .Actually , I have 15,000,000 count functions in the
231 wbooks ;
here begin another story , how I can built them ....I find a method
after months
of searchings , with an VBA code , which allow you to built 65536
Count functions
(with the 15 references in combinatoric order! ) in some few
minutes .Here is what I
have actually .

Normally , I look and search for speed ; a bigger speed mean a bigger
database !
with the new formula I have posted here at the begin of this thread ,
I only need 4
or 5 columns of this kind of functions in every wbook , and no more
autofill method ;
in this wey , VBA macro only open the wbook , loops through an AND
function placed in Column
BD 91:BD65536 , and if the value of cell is TRUE , copy antire row and
paste it in another
wbook ! In this wey , with the same resource sistem , I'll have a
database of 5 size bigger ;

And now I get to the reason of this thread , why I need an equivalent
for this function , (if even
it works perfect for me ! ) . If i have an equivalent for this
function , in a kind like this , for eg. :
=COUNT(INDEX(MATCH(ROW(B4:AH83),
{4,7,9,11,14,16,18,22,25,35,46,57,68,72,83},0)/
ISNUMBER(B4:AH83),0)) ,

... I,ll can built this functions more easyer with this method I
found ( with that VBA
macro ! ...) . To understand perfect I must send you some wbooks to
see clarly
how I built this functions , and I can send you them .The best and
fastest wey I
think to be to give you my messenger ID , there I can send you the
wbooks and
explain instantly .My messenger ID is (e-mail address removed) , if you
agree !

Thanks very much for your time !