AND function

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ytayta555 - 12 May 2008 19:30 GMT

A good day !

I have this function :

=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) ;
this function let me to know if every one of the 5 COUNT functions
have the result the value <=1;
(this is what I need !!, not to count everyone of the five count
functions the value <=1!!!)

Can you help me to get a function which do the same thing like this
function ?
I get some variants , but them doesn'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))
or

=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

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

Thanks very much for your time

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Tiago Nogueira - 12 May 2008 19:55 GMT

Try to use the "OR" fstion instead of "AND"...

it may work.

> A good day !
>
[quoted text clipped - 24 lines]
>
> Thanks very much for your time

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ytayta555 - 12 May 2008 20:05 GMT

On 12 Mai, 21:55, Tiago Nogueira
<TiagoNogue...@discussions.microsoft.com> wrote:

> Try to use the "OR" fstion instead of "AND"...
>
> it may work.

Thank for sugestion ! First formula (with
AND....=AND(COUNT(B1;B3;B5)<=1;COUNT(C1....)
work perfectly for me ! What I need very strong is another function
which to do exactly the
same thing for me !

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Bob Phillips - 12 May 2008 20:15 GMT

Why? If it works, why do you need another?

Signature

---
HTH

Bob

(there's no email, no snail mail, but somewhere should be gmail in my addy)

> On 12 Mai, 21:55, Tiago Nogueira
> <TiagoNogue...@discussions.microsoft.com> wrote:
[quoted text clipped - 7 lines]
> which to do exactly the
> same thing for me !

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ytayta555 - 12 May 2008 20:41 GMT

> Why? If it works, why do you need another?

THANKS!
OH ! It's a long story !...How can I explain ? ...I'll try !

I was looking for many months to find a method to do Count functions
with references in combinatoric order , such as : COUNT(C1;C2;C3)
, then COUNT(C1;C2;C4),COUNT(C1;C2;C5),COUNT(C1;C3;C4) ...
etc ...etc ..; For my need I must have 15 references ; I has need
thousand
and thousand of this functions ; Now , I have 15.000.000 Count
functions with
the arguments/references in combinatoric order , so , no one of them
is not
identical with another ! I find a method which is doing 65536 Count
functions
with the arguments/references in combinatoric order in some few
minutes !
This thing , with a combination of VBA code and excel ;

Now , I need another function , and I try (and I can do this thing
now) to built it
in the same wey . But , if I'll get another variant for my function ,
I'm shure I'l
built it more easy .

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ytayta555 - 12 May 2008 20:51 GMT

> Why? If it works, why do you need another?

Sir Bob Philips , do you want to help me , and to give me an e-mail
or gmail adress to can explain you better my problem ?
(and , if you have need , how I find to built this kind of
Count functions ?) What for are helpfull this kind of functions ??
This kind of functions are very usefull for persons who have
lotto statistics hobby ! I was looking for to find this method many
months ...The method can be applicate and to another functions
similary with Count function .
Thanks for your time

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ytayta555 - 12 May 2008 21:29 GMT

Sir T. Valko , do you want to help me , please ?
Many thanks

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