SQL Server 2019

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select concat(a,b,c,d)
from (select 'a','b','c','d') as t(a,b,c,d)
group by cube(a,b,c,d)
having len(concat(a,b,c,d)) = 3


(No column name)


bcd

acd

abd

abc


select coalesce(a, 0) + coalesce(b, 0) + coalesce(c, 0) + coalesce(d, 0)
from (values (1, 2, 3, 4)) t(a,b,c,d)
group by cube(a,b,c,d)
having (case when a is not null then 1 else 0 end +
        case when b is not null then 1 else 0 end +
        case when c is not null then 1 else 0 end +
        case when d is not null then 1 else 0 end
       ) = 3


(No column name)


9

8

7

6


with t as (
      select t.*
      from (values (1), (2), (3), (4)) t(x)
     )
select t1.x + t2.x + t3.x
from t t1 join
     t t2
     on t1.x < t2.x join
     t t3
     on t2.x < t3.x


(No column name)


6

7

8

9