Frequency of Pairs Formula

[srizki]

Hi all,
I have cost centers in column A, and corresponding account codes in column B. I want to know how many different pairs are there, for each cost center and accounting codes.

So something like,
=IF(SUMPRODUCT(($A$1:A1000=A1)*($B$1:B1000)),…

Thanks

 

[Marcelo Branco]

Could you provide a small data sample (~10 rows) along with expected results?

M.

 

[srizki]

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[TD]88001610
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[TD]411023314
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[TD]88001610
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[TD]411023314
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[TD]88001610
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[TD]411023314
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[TD]88001610
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[TD]411023314
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[TD]88001610
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[TD]411023314
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[TD]88001610
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[TD]411023314
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[TD]88001610
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[TR]
[TD]411023314
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[TD]88001610
[/TD]
[/TR]
[TR]
[TD]411023314
[/TD]
[TD]88001610
[/TD]
[/TR]
[TR]
[TD]411023314
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[TD]88001610
[/TD]
[/TR]
[TR]
[TD]411023314
[/TD]
[TD]88001610
[/TD]
[/TR]
[TR]
[TD]411023314
[/TD]
[TD]88001610
[/TD]
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[TR]
[TD]411023314
[/TD]
[TD]88001610
[/TD]
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[TD]411023314
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[TD]88300110
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[TD]411023314
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[TD]88500100
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[TD]411039133
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[TD]88001610
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[TR]
[TD]411039133
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[TD]88001610
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[TD]411039133
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[TD]88001610
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</tbody>[/TABLE]

So what I am looking for is, how many times each cost center with each accounting code. Each cost center may have many accounting codes and each accounting code may be with more than one cost center.
I would like to know the combinations.
Something like your earlier post.
=COUNTIFS(A1:A5,"Health",B1:B5,1)

 

[Gerald Higgins]

So with that data, what should the answer be ?

 

[srizki]

I will be 16
1
1
3

I used Marcelo's formula
=COUNTIFS(A1:A5,"Health",B1:B5,1),
and it is working,
I think that will suffice.

Thanks for reading, really appreciated, both of you have helped me in the past.