Prakash, Just in case I'm still misinterpeting the problem, here's an approach that uses your original data set as input.  It creates an observation for each 72C3 combination that have all 3 items equal to 1. data _72C3_ (keep=_72C3_);    set original;    array items {72} item_1 - item_72;    do i = 1 to 70;       do j = i+1 to 71;          do k = j + 1 to 72;              if items{i} = items{j} = items{k} = 1 then do;                 _72C3_ = put(i, z2.) || ' ' || put(j, z2.) || ' ' || put(k, z2.);                 output;             end;          end;       end;    end; run; Then get counts for each value of _72C3_. Good luck. ... View more