hi bkolibov/gurpinder:
an official counting problem that sets a classic trap in the final step. gurpinder's work is perfect until the end where he makes the mistake of adding instead of multiplying. this is one of the most common traps that I have seen used by the test makers for these types of combinations.
first determine how many unique combinations of senior officers are possible: 6C3 = 20 (as gurpinder calculated)
second determine how many unique combinations of junior officers are possible: 4C1 = 4 (as gurpinder calculated)
at this point you must multiply to get the correct answer of 80. Why? Remember the counting tree. The only time you add results in counting problems is when you are combining separate counts in problems with wording such as "at least", "at most" or "or" and on some restriction problems where you are adding up allowed cases. Imagine that the 20 possibilities for senior officers are labeled with the numbers 1-20 and the 4 possibilities for junior officers labeled A,B,C,D. Here is a sampling of the 80 possibilities:
1A, 1B, 1C, 1D, 2A, 2B, 2C, 2D, 3A, 3B..........20A, 20B, 20C, and 20D 80 unique groups in total. Hope this helps and let me know if I can clarify futher!
