Function Equivalent need
 

HI ALL , and a good day

I have a very big need ! I have this
combined function :

=AND(COUNT(B1;B3;B5)<=1;COUNT(C1;C3;C3)<=1;COUNT(D 1;D3;D5)<=1;COUNT(E1;E3;E5)<=1;COUNT(F1;F3;F5)<=1)


what I need is a function to be equivalent (!) to this (!) function
posted here

any sugestion will be more then wellcome

Function Equivalent need
 

How about

=COUNTIF(B1:F1,"<=1")+COUNTIF(B3:F3,"<=1")+COUNTIF (B5:F5,"<=1")

Hope this helps,

Hutch

"ytayta555" wrote:

HI ALL , and a good day

I have a very big need ! I have this
combined function :

=AND(COUNT(B1;B3;B5)<=1;COUNT(C1;C3;C3)<=1;COUNT(D 1;D3;D5)<=1;COUNT(E1;E3;E5)<=1;COUNT(F1;F3;F5)<=1)


what I need is a function to be equivalent (!) to this (!) function
posted here

any sugestion will be more then wellcome

Function Equivalent need
 

Does this work for you:

=COUNT(B1:F1;B3:F3;B5:F5)<=1

?
--

HTH,

RD
================================================== ===
Please keep all correspondence within the Group, so all may benefit!
================================================== ===

"ytayta555" wrote in message
...
HI ALL , and a good day

I have a very big need ! I have this
combined function :

=AND(COUNT(B1;B3;B5)<=1;COUNT(C1;C3;C3)<=1;COUNT(D 1;D3;D5)<=1;COUNT(E1;E3;E5)<=1;COUNT(F1;F3;F5)<=1)


what I need is a function to be equivalent (!) to this (!) function
posted here

any sugestion will be more then wellcome

Function Equivalent need
 

On 6 Mai, 17:30, Tom Hutchins
wrote:
How about
=COUNTIF(B1:F1,"<=1")+COUNTIF(B3:F3,"<=1")+COUNTIF (B5:F5,"<=1")


Thanks very much , Tom

It is indeed a variant for me ; please to help me with another
variant ; I
have get this formula :

=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))

Can you make ajustaments and match with this formula
I get ,for my function posted here ?

Thanks very much for your time

Function Equivalent need
 

Thanks RagDyeR

Your variant doesn't work for me because it don't
show me if every of 5 COUNT functions have
the value <=1 ;in my first example , AND function
show me if every of 5 Count functions have value
= ,,<=1,,

Function Equivalent need
 

I'm not sure what you want...

To adapt the SUMPRODUCT formula to give the same results as our earlier
multiple COUNTIF, use

=SUMPRODUCT(ISNUMBER(MATCH(ROW(B1:F5), {1,3,5},0))*(B1:F5<=1))

Is that what you wanted?

Hutch

"ytayta555" wrote:

On 6 Mai, 17:30, Tom Hutchins
wrote:
How about
=COUNTIF(B1:F1,"<=1")+COUNTIF(B3:F3,"<=1")+COUNTIF (B5:F5,"<=1")


Thanks very much , Tom

It is indeed a variant for me ; please to help me with another
variant ; I
have get this formula :

=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))

Can you make ajustaments and match with this formula
I get ,for my function posted here ?

Thanks very much for your time

Function Equivalent need
 

Thanks very much , mister Tom Hutchins
This help me really ; I understand now that I must
do my homework with this four functions (SUMPRODUCT,
ISNUMBER,MATCH,ROW), to get finally what I need ;
excuse me for late feedback .

I'll post today or tomorrow what I try to do ,
because of miss of time now ;
I really got a new way for me ; hope to visite me
here after I'll post with details what I try to do ;

I think shall be something very usefull for a lot
of excel users , I find after some few months how
to make 65536 Count functions with the arguments(references)
in combinatoric order , in a few minutes ;
such as :
COUNT(B1;B2;B3;B4)
COUNT(B1;B2;B3;B5)
COUNT(B1;B2;B3;B6)
COUNT(B1;B2;B3;B7)
COUNT(B1;B2;B4;B5) etc..etc...to 65536 functions.....
How many time take anybody to make 65536 functions
like this ??? But , if you need (like me) to have 15 arguments/
references in
a only Count function , how long time take ???
I-ll explain what solution I have found , and what I need actually .

Thanks for your time