Case 1: The committee consists of 3 seniors and 2 juniors. Case 2: The committee consists of 4 seniors and 1 junior. Case 3: The committee consists of 5 seniors and 0 juniors. 7C3∙5C2 + 7C4∙5C1 + 7C5∙5C0 = 35∙10 + 35∙5 + 21∙1 = 350 + 175 + 21 = 546 ways Edwin