To solve for the different ways the gift certificates will be awarded:
Using the Basic Counting Rule, there are 34 choices for the first person, 33 remaining choices for the second person and 32 for the third, so there are 34 (33) (32) = 35,904 ways to choose three people. Suppose for a moment that Abe is chosen first, Bea second and Cindy third; this is one of the 35,904 possible outcomes. Another way to award the prizes would be to choose Abe first, Cindy second and Bea third; this is another of the 35, 904 possible outcomes. But either way Abe, Bea and Cindy each get $50, so it doesn’t really matter the order in which we select them. In how many different orders can Abe, Bea and Cindy be selected? It turns out there are 6:
ABC ACB BAC BCA CAB CBA
How can we be sure that we have counted them all? We are really just choosing 3 people out of 3, so there are 3 × 2 × 1 = 6 ways to do this; we didn’t really need to list them all, we can just use permutations!
So, out of the 35, 904 ways to select 3 people out of 34, six of them involve Abe, Bea and Cindy. The same argument works for any other group of three people (say Abe, Bea and David or Frank, Gloria and Hildy) so each three-person group is counted
six times. Thus the 35, 904 figure is six times too big. The number of distinct three-person groups will be 35, 904/6 = 5984.
Hence the different ways the gift can be awarded is 5984 ways