Counting frequency based on Unique values in 2 columns

[Brunts]

Here's the type of data I'm dealing with. I know that it would be considerably easier if it was one row per defect, but each row on the defect has other unique information in it (That I don't need for the summary) but do for other purposes so can't delete it)

Column A	Column B	Column C	Column D	Column E	Column F	Column G	Column H
Row 1
Row 2		Defect No	Defect Outcome		No of Defects
Row 3		1	Pass		Pass	2
Row 4		1	Pass		(blank)	1
Row 5		1	Pass
Row 6		1	Pass
Row 7		1	Pass
Row 8		1	Pass
Row 9		2
Row 10		2
Row 11		2
Row 12		2
Row 13		2
Row 14		3	Pass
Row 15		3	Pass
Row 16		3	Pass
Row 17		3	Pass
Row 18		3	Pass

<tbody>
</tbody><colgroup><col><col span="2"><col><col><col><col><col></colgroup>

The values in Column D will always be the same for each Unique Defect No (Except in error, but no need to cater for!) and I have no idea how many Defects there will be, so presume a range of up to 500 will suffice.
What I'm after is a formula for Cell G3 that will confirm a count of Unique Defect Nos that have the pass criteria or blank value. Been trawling 'internet and having no luck with sumproducts, countif, array formulas etc.

thx

 

[AhoyNC]

This may do what you want.
Will the defect always be all of one number passes or fails?
With the formulas below if row 10 had a pass and the other rows for defect# 2 are blank it will show 3 pass and 1 fail.

These are array formulas and must be entered with CTRL-SHIFT-ENTER.
If your defect numbers are numeric and not text you could use the formula in I3 & I4.

Excel Workbook
	A	B	C	D	E	F	G	H	I
1	Row 1
2	Row 2		Defect No	Defect Outcome		No of Defects
3	Row 3		1	Pass		Pass	2		2
4	Row 4		1	Pass		Blank	1		1
5	Row 5		1	Pass
6	Row 6		1	Pass
7	Row 7		1	Pass
8	Row 8		1	Pass
9	Row 9		2
10	Row 10		2
11	Row 11		2
12	Row 12		2
13	Row 13		2
14	Row 14		3	Pass
15	Row 15		3	Pass
16	Row 16		3	Pass
17	Row 17		3	Pass
18	Row 18		3	Pass
Sheet

 

[Brunts]

That certainly explains why I couldn't work it out - works a treat - well done you

 

[AhoyNC]

You're welcome. Thanks for the feedback.