Case 1: When we have 3 coffee cups in one row, 2 coffee cups in another row, and 1 coffee cup in the remaining row (3!). For row having 3 coffee cups arrangement can be done in only one way. For row having 2 coffee cups arrangement can be done in three ways. For row having 1 coffee cup arrangement can be done in three ways. Total arrangements for case 1=3!×1×3×3=54. Case 2: When we have 2 coffee cups in each row (1). Each row has 2 coffee cups arrangement can be done in three ways. Total arrangements for case 2=1×3×3×3=27 So, total arrangement for given conditions =54+27=81.