Practice Problem - Lesson 5
Could someone clarify the example problem at the end of lesson 5 that is labeled "solution"? It begins "Nine people will be selected from a group of 20 for a photograph." I'm not understanding why for Set A, the total number of available people to choose from is 17, when we are choosing at most 1 of the twins and at most 1 of the triplets out of 20. Similarly, for Set B, there are 16 people left to choose from, but we chose both twins and at most 1 triplet out of 20. Why is this?
katrina: Jan. 28, 2015, 2:02 a.m.
For set D, why is it P(15,5) if both twins are chosen and two triplets are chosen? Shouldn't it be P(16,5)?
martha2A: Jan. 30, 2015, 12:36 a.m.
for set A the total number of people chosen is 17 because we want at most 1 twin and at most 1 triplet so we remove the two triplets and one twin that aren't going to be chosen in order to only leave the possibility for at least one twin and at least one triplet to be chosen.
martha2A: Jan. 30, 2015, 12:41 a.m.
I am not sure if my logic is correct for this problem though.
weisbart: Feb. 1, 2015, 10:58 a.m.
Once again, you have it right martha2A, and with a nice explanation. We have only 17 to choose from since we have to throw out two triplets and one twin--We can only chose at most one of each. \[\] For katrina's question, since two triplets are chosen, we have to throw out all three. Why? Because if you left the third in, you'd have the opportunity to choose him. So both twins and all triplets must be thrown out.