Arguments Count Function problem

Y

ytayta555

Hi everybody

I have 33 Count functions :

=COUNT(B46;B47;B48;B49;B50;B51;B52;B53;B54;B55;B56;B57;B58;B62;B63),
=COUNT(C46;C47;C48;C49;C50;C51;C52;C53;C54;C55;C56;C57;C58;C62;C63),
.....................
=COUNT(AH46;AH47;AH48;AH49;AH50;AH51;AH52;AH53;AH54;AH55;AH56;AH57;AH58;AH62;AH63)

How must look this function to have the same rows number , but to
count 33
columns , from B to AH ??
This 33 functions , can be in only one ??

Please very much to help !
 
R

Ron Coderre

Try this:

=COUNT(B46:C58,B62:C63,AH46:AH58,AH62:AH63)

Is that something you can work with?
Post back if you have more questions.
--------------------------

Regards,

Ron
Microsoft MVP (Excel)
(XL2003, Win XP)
 
R

Ron Coderre

Amended formula:
=COUNT(B46:AH58,B62:AH63)

Does that help?

Regards,

Ron
Microsoft MVP (Excel)
(XL2003, Win XP)
 
Y

ytayta555

Thank very much

I must have 15 arguments , like in :
=COUNT(B46;B47;B48;B49;B50;B51;B52;B53;B54;B55;B56;B57;B58;B62;B63)
but to contain in only one formula (it can be and imbricate) all this
33 columns :
b,c,d,e,f..........ah ! And the rows to be the same ! But it MUST to
have 15
arguments !

Respectfully
 
Y

ytayta555

I have a big problem !
I have many functions which have the arguments in this way ,
foe example :
=COUNT(B4;B7;B9;B11;B14;B16;B18;B22;B25;B35;B46;B57;B68;B72;B83)
I mean the rows are not in continuous mode ;
What I can do in this case ? Please for help !

I have a database of 231 with 15,000,000 count functions
with the references in combinatoric order ! I use it for ....
lotto statistic .
Many thanks again
 
Y

ytayta555

.....231 workbooks(I mean) ! By the wey ! Know somebody how
you can do a so big number of functions ??
If somebody
try to create this functions with the references {15!} in
combinatoric order , it shall understand how hard is
it ! I was looking to resolve this problem a few of
months , and I found a solution .
 
R

Ron Coderre

It's really important that you present your exact structure and what you
want to do in your first post. Otherwise, you could have many talented
people wasting their time solving a problem you don't have.

That being said...
try one of these:
=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))

or
=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))

Does that help?
--------------------------

Regards,

Ron
Microsoft MVP (Excel)
(XL2003, Win XP)
 

Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments. After that, you can post your question and our members will help you out.

Ask a Question

Similar Threads


Top