Поиск наиболее часто встречающегося подмножества с помощью Select

declare @table  table(id int,letter char(1))
insert into @table (id ,letter)
select id ,letter
from
(
select 1 as id,'a' as letter
union
select 1,'b'
union
select 1,'c'
union
select 2,'a'
union
select 2,'b'
union
select 2,'c'
union
select 3,'a'
union
select 3,'c'
 
) as t
select count(t.l1+t.l2),(t.l1+t.l2)
from
(
select t.id,t.letter as l1,t1.letter as l2 
from @table as t  join @table as t1 on t.id=t1.id and not t.letter=t1.letter
where not t.letter=t1.letter
) as t
group by (t.l1+t.l2)

declare @table  table(id int,letter char(1))
declare @table1  table(l1 varchar(10))
declare @table2  table(id int,l1 varchar(10))
declare @result  table(comb varchar(10), qty int)
declare @id int
DECLARE @var varchar(8000)--комбинации letter
insert into @table (id ,letter)
select id ,letter
from
(
select 1 as id,'a' as letter union select 1,'b' union select 1,'c' /*union select 2,'a'*/ union select 2,'b' union select 2,'c' union select 3,'a' union select 3,'b' union select 3,'c'
) as t
select *
from @table
--цикл по id
set @id=1
while exists (select distinct id from @table where id=@id)
begin
    --добавляем комбинацию letter для очередного id
    insert into @table1 (l1)
    select letter
    from
    (
    select  letter
    from @table where id=@id
    ) as t 
--  select *
--  from @table1
    SET @var = ''
    SELECT @var = @var + l1 FROM @table1 
--  SELECT @var
    insert into @table2 (id,l1)
    values( @id,@var)
    set @id=@id+1
    delete @table1
end
    select *
    from @table2
 
 
declare @q as int
declare @p as varchar(10)
declare @p1 as varchar(10)
set @p=(select top 1 l1 from @table2 order by len(l1) desc )
select @p
set @q=len(@p)+1
--проходим @p в прямом порядке
while @q>0
begin
    set @p1='%'+isnull(stuff(@p,@q,1,'%'),@p)+'%'
insert into @result (comb, qty )
select @p1, count(patindex(@p1,l1)) as qty from @table2
where patindex(@p1,l1)>0
set @q=@q-1
end
 
select * from @result