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Padma, I did the traditional way, lets say 1,2,3,4,5,Ben and Ann are 7 contestants. First Ben and Ann are ruled out, we r left with 1,2,3,4,5 and so the possible selections are 1234, 1235, 1245,1345 and 2345.
Well I have the innovative way. I hate to depart from classical way to solve this.
Let us say we have chosen A & B now now we choose 2 out of remaining 5
we get a pair and let us call it C
so we have A,B,C we can arrange these in 3 ways
(A,B,C), (C,A,B), (B,C,A)
so we have 3 * 5C2 combinations in which A and B are present = 30
Total combinations = 7C4 = 35