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;AH5 4;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 !

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)

"ytayta555" wrote in message
...
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;AH5 4;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 !

May be

=COUNT(B46:H63)


"ytayta555" wrote in message
...
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;AH5 4;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 !

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

Does that help?

Regards,

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

"Ron Coderre" wrote in message
...
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)

"ytayta555" wrote in message
...
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;AH5 4;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 !

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

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;B5 7;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

.....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 .

And in excel 2003 you can not use more of
30 arguments ! .......

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)


"ytayta555" wrote in message
...
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;B5 7;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