Hi binaras,

This is a rare layered Combinatorics question. The most efficient way to answer it is to use subtract the options that you "don't want" from the total possible options (the approach that Mitch used); you could also calculate what you DO want, but that requires 3 separate multi-step calculations...

We're given 4 senior partners and 6 junior partners. We're asked for the number of different groups of 3 (the clue that we'll need the Combination Formula) with one stipulation - there must be AT LEAST 1 senior partner. Here are the 3 calculations:

1) 3 senior partners = 4c3 = 4!/[3!1!] = 4 different groups

2) 2 seniors and 1 junior = (4c2)(6c1) = (4!/[2!2!])(6!/[1!5!) = (6)(6) = 36 different groups

3) 1 senior and 2 juniors = (4c1)(6c2) = (4!/[3!1!])(6!/[2!4!]) = (4)(15) = 60 different groups

4 + 36 + 60 = 100 different groups

Final Answer: B

GMAT assassins aren't born, they're made,
Rich