Thanks, but I am already aware of this formula. There was a previous post
where this solution was proposed. What I really wanted was a way of getting
the same result using SUMPRODUCT, but I am afraid it might not be possible.

Regards,

Paulo

"T. Valko" wrote:

Try this array formula** :

=SUM(IF(FREQUENCY(IF(All!M2:M5514=A22,MATCH(All!B2 :B5514,All!B2:B5514,0)),ROW(All!B2:B5514)-ROW(All!B2)+1),1))

** array formulas need to be entered using the key combination of
CTRL,SHIFT,ENTER (not just ENTER). Hold down both the CTRL key and the SHIFT
key then hit ENTER.

--
Biff
Microsoft Excel MVP


"Paulo" wrote in message
...
No, but then the first part of the function (all!$B$2:$B$5514<"") would
have
taken care of that.

Let's assume there are no empty cells, if this helps.

Thanks,

Paulo


"T. Valko" wrote:

Are there any empty cells within the range all!B2:B5514?

--
Biff
Microsoft Excel MVP


"Paulo" wrote in message
...
This question was posted before but the solution proposed used
different
functions. I would like to insist. Is there a modification to the
formula
below, using SUMPRODUCT, that allows to retrieve the number of unique
itens
that meet a 2nd criteria?

Original formula:
=SUMPRODUCT((A1:A20<"")/COUNTIF(A1:A20,A1:A20&""))

In my case, I need the numbers of unique branches (column B) that meet
a
specifc criteria in another column (M). So, I have the formula:

=SUMPRODUCT((all!$B$2:$B$5514<"")/COUNTIF(all!$B$2:$B$5514,all!$B$2:$B$5514&""))

but I need to include the 2nd criteria:

(all!$M$2:$M$5514=$A22)


Thanks in advance,

Paulo




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