Count the number of different values based on MULTIPLE columns

Thank you a lot for your perusal, Brian. Just edited my post.

pleasure Tho,
my assumption is you would like to count something based on a certain criteria?
please clarify.
apologies for delayed uptake, its been a long day but im happy to review

I didnt even expect this quick replies from the community. It's so kind of you Brian.
I did edit my post. I will put the calarification here again.

For clarification, the dataset is like this:
A b c d
province, district, village, Households
p1 d1 v1 h11

p2 d2 v2 h21
p2 d2 v2 h21

p3 d3 v3 h31
p3 d3 v3 h31

p4 d41 v41 h41
p4 d41 v41 h42

p4 d42 v42 h43

Then there are only 5 triplets of (p,d,v)={(1,1,1),(2,2,2),(3,3,3),(4,41,41),(4,41,42)}
There are some duplicates of (p,d,v).

Yes, Brian. I want to count the number of distinct triplets (p,d,v). It's similar to "the households in each village dont matter".

Donno if I have made it clear enough.

sorry tho,
would have loved to help but i think a sample file would be useful
anyway i know that to do this you would require nesting the countifs in a sumproduct..i have this but tyou would have to edit to suit

=SUMPRODUCT((1/COUNTIFS(A2:A22,A2:A22,B2:B22,B2:B22,C2:C22,C2:C22)))

i hope this helps

Hello Brian,

I traced through your instructions, just made it now. Thank you so much, Brian. You are a life saver.

Brian,

It's very similar to yours. Took time to understand what you wrote :D
=SUMPRODUCT(1/COUNTIFS(A2:A9223,A2:A9223,B2:B9223,B2:B9223,C2:C9223,C2:C9223,D2:D9223,D2:D9223,E2:E9223,E2:E9223))

Just add reputation to your account and will mark it as solved now.

Yea, it messes with notions of reality this one, you doubt its practicality the whole time youre building it until you understand it.
As a side note, it is quite slow as it is doing a scan of everything, there is an alternative way using frequency nested in a sum product, which does it in half the time or less over huge amount of data, so if speed may be a factor for you or size of d set, I recommend explore that.
Thanks for the rep and prompt response.
Good luck.