Bunuel wrote:
Jane and Thomas are among the 8 people from which a committee of 4 people is to be selected. How many different possible committees of 4 people can be selected from these 8 people if at least one of either Jane or Thomas is to be selected?
A) 28
B) 46
C) 55
D) 63
E) 70
Kudos for a correct solution.
Number of ways of selecting 'r' from total 'n' is given by formula ncr = n!/(r!(n-r)!)
Oder is not important in forming a committee.
Number of ways if at least one of Jane or Thomas is selected = Number of ways (Jane only is selected + Thomas only is selected + both Jane and Thomas are selected)
So, if Jane is selected in the committee, number of ways other 3 members are selected from group of 6 (8 - Jane and Thomas)
(J _ _ _)= 6C3 = 6!/(3!*(6-3)!) = 20 ways
if Thomas is selected in the committee, number of ways other 3 members are selected from group of 6 (8 - Jane and Thomas)
(T _ _ _) = 6C3 = 6!/(3!*(6-3)!) = 20 ways
if both Jane and Thomas are selected in the committee, number of ways other 2 members are selected from group of 6 (8 - Jane and Thomas)
(T J _ _) = 6C2 = 6!/(2!*(6-2)!) = 15 ways
Adding all .. 20+20+15 = 55 ways.
Answer (C)