Every once in a while, I stand back in amazement at solutions that end up
fixing a problem. This is one of those times.

Generally, I research the suggestions, learn something new and apply it to
my problem. Thanks to this group, I have been able to develop some pretty
neat spreadsheets that solve problems faced by the small group of people I
work with.

This solution, while it seems to work beautifully, I cannot, for the life of
me, figure out why or how it works.
Upon review, I viewed SUMPRODUCT as a shortcut to lots of multiplying and
adding, and saw little of value and could not understand how it applied to
my problem. I have learned to never disregard advice given in this forum and
therefore, simply pasted your formula in my spreadsheet with the necessary
address changes to fit, it worked. Why? I have no earthly idea!

Specific questions: What does the -- mean to the formula? What does the
<"" do? and Since I thought that SUMPRODUCT multiplyed "Group A" against
"Group B" then added the answers for a single sum. The Data in the Range
given in the formula is text and I thought that SUMPRODUCT would treat it as
a ZERO.

With all this said: IT WORKED!

Is there a source of wisdom somewhere online that would help me out?

In your debt,
Craig



"Peo Sjoblom" wrote in message
...
Or use something else than COUNTA

=SUMPRODUCT(--(B2:B20<""))


--

Regards,

Peo Sjoblom


"Ken Johnson" wrote in message
ups.com...
On Aug 11, 7:33 am, "C Brandt" wrote:
I ran across an interesting problem. I need to count the number of
valid
entries and I simply used an IF statement to apply criteria to a column
of
cells and copied only those I was interested in counting to the
adjacent
column, then used COUNTA( ) to count them.
B12 is equal to =IF(and(A12 <MAX,A12MIN,A12,"")
Didn't work. It counts the "" cells.
If I manually delete one of the "" entries in column B, it doesn't
count
it.
Is there a simple solution?

Thanks,

Craig

Hi Craig,

One way would be to stop using "" and use something like "NO" instead,
then use COUNTIF(B:B,"NO").

Ken Johnson