gurudabl wrote:
Hi
ScottTargetTestPrep Could you help me with my query?
Quote:
If none of the candidates is eligible for a position in both departments
What role does this phrase play in the question? Does it affect our calculation at all?
Thank you,
Dablu
The fact that none of the candidates is eligible for both departments play a significant role in the question. To compare, suppose that there was one person who is eligible for a position in both departments. First of all, since that one person is included both in the 7 candidates for the math department and 10 candidates for the CS department, there are no longer 17 total candidates; there are only 7 + 10 - 1 = 16 total candidates. Further, the number of sets of 3 candidates change as well. For the math position, we still have 7 choices; however, the number of ways we can fill the CS position depends on whether the person filling the math position is the one eligible for both positions or not. If it is, then there are only 9 people you can choose from to fill the CS position; so in this scenario, the number of ways to choose 2 people from a total of 9 people is 9C2 = 9!/(2!*7!) = (9*8)/2 = 36. On the other hand, if the person chosen for the math department is one of the remaining six people, then we can fill the CS positions by choosing 2 people from a total of 10 people, which can be done in 10C2 = 45 ways, as calculated before. Thus, when there's one person eligible for both positions, the number of ways to fill the three positions is 36 + 6*45 = 306.
If you're curios what happened to the 315 - 306 = 9 ways to fill the three positions, let X be the person eligible for both positions. Since X is included in both the 7 and 10 people, the choices X + XA, X + XB, ... are no longer possible, where A, B, ... denote the people eligible for the CS department besides X. As there are 9 people eligible for the CS department besides X, the total number of ways is reduced by 9.