On 4 Iun, 13:19, Joel wrote:
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 , if you
agree !

Thanks very much for your time !