Access - Get distinct count of rows based on different column values

The easiest way to create a transform query like this and save it

TRANSFORM Count(YourTable.ID) AS CountOfID
SELECT YourTable.ID
FROM YourTable
GROUP BY YourTable.ID
PIVOT YourTable.Period)

Which outputs

ID  1 2 3
--  - - -
13  1 1 
25  1 1 1
52  1   1
94  1 1 1

You can then use that create another query Lke this

SELECT t.Groups, Count(t.ID) AS CountOfID
FROM (SELECT ID, 
     SWITCH(
               [1] = 1 and [2] =1 and [3] = 1, 'Periods 1, 2, and 3',
               [1] = 1 and [2] =1 , 'Periods 1 and 2',
               [1] = 1 and [3] =1 , 'Periods 1 and 3'
               )  as Groups

FROM TheSavedTransformQuery)  AS t
GROUP BY t.Groups;

Which outputs this

Groups              CountOfID
------------------- ---------
Periods 1 and 2        1
Periods 1 and 3        1
Periods 1, 2, and 3    2

If you want to do it as single SQL statement this will work as well

SELECT 
    t.Groups, 
    Count(t.ID) AS CountOfID
FROM (

    SELECT 
        Switch(
            FirstPeriod.Period=1 And [SecondPeriod.Period ]=2 And [ThirdPeriod.Period]=3,'Periods 1, 2, and 3',
            FirstPeriod.Period=1 And [SecondPeriod.Period]=2,'Periods 1 and 2',
            FirstPeriod.Period=1 And [ThirdPeriod.Period]=3,'Periods 1 and 3') AS Groups, 
            DistinctID.ID
    FROM (((SELECT DISTINCT ID
        FROM 
        YourTable )  AS DistinctID 

        LEFT JOIN (SELECT ID, PERIOD 
        FROM YourTable
        WHERE Period = 1)  AS FirstPeriod 

        ON DistinctID.ID = FirstPeriod.ID) 

        LEFT JOIN (SELECT ID, PERIOD 
        FROM YourTable  WHERE Period = 2)  AS SecondPeriod 
        ON DistinctID.ID = SecondPeriod.ID) 

        LEFT JOIN (SELECT ID, PERIOD 
        FROM YourTable
        WHERE Period = 3)  AS ThirdPeriod 
        ON DistinctID.ID = ThirdPeriod.ID

)AS t
GROUP BY 
    t.Groups