A.) If there are no restrictions, Number of committes possible = 9C5 = 126
b.) If both Jim and Mary must be in the committee, then we are now choosing 3 persons from the remaining 7 people. So the number of committes possible = 7C3 = 35
c.) If either Jim or Mary (but not both) must be on the committe, then there are 7 people availabe and 4 choices to make after taking either Jim or Mary. Also, there are 2 possible ways of choosing either Jim or May. Theefore, Number of committe possible = 2(7C4) = 2(35) = 70
