Factorials? to calculate values for multiple group combinations

[yomero]

I have 4 groups {A, B, C, D}, in a column called "Event" .

How in MS Access can I calculate the unique/distinct combinations.
Is there a way to do it without creating individual queries per group?

For example:

Groups
A
A+B
A+B+C
A+B+C+D
A+C
A+C+D
A+D
B
B+C
B+C+D
B+D
C
C+D
D

How can I create the groups in MS Access (instead of using many queries)

 

[alansidman]

Create a Group By Query. Here is an example that counts the number of unique items (field1)

SELECT Sheet1.field1, Count(Sheet1.field1) AS CountOffield1
FROM Sheet1
GROUP BY Sheet1.field1;

 

[Micron]

Create a Group By Query. Here is an example that counts the number of unique items (field1)

SELECT Sheet1.field1, Count(Sheet1.field1) AS CountOffield1
FROM Sheet1
GROUP BY Sheet1.field1;

Won't this simply return the number of times A, B, C and D are found in the column?

I'm guessing you want the possible combinations of a population taken one at a time (since you started only with A) and not just the numer. Excel has a native function (Combin, I think) but it will only give you the number. I've read of other ways to do this, but have never tested any of them.
If your data is in one table only, try creating a query with the table in 2x and don't create any joins. Something like
SELECT A, B, C, D FROM tbl1, tbl2;
You should get the Cartesian product and can evaluate the number of records using the Excel function to cross-check the result.

 

[yomero]

Yes Micron, I need all the combinations which I will use to calculate the values (not count) for each of them.

Thanks. I will try this approach.

 

[alansidman]

Won't this simply return the number of times A, B, C and D are found in the column?

Yes. That was my understanding of the the problem. Apologies if not what you were looking for.

 

[Micron]

Yes. That was my understanding of the the problem. Apologies if not what you were looking for.

I'd say no apologies required, but I'm like you in that respect. Funny how we should be so inclined when all we're trying to do is help. Keep on keepin on! You've been at it a lot longer that some of us, I see!

 

[yomero]

Yes. That was my understanding of the the problem. Apologies if not what you were looking for.

I appreciate your time for replying AlanSidman.
No worries.