Mr. Green teaches band, choir, and math. This year, he has 57 students that take at least one of his classes. He teaches band to 25 students. There are 9 students who have Mr. Green for math and nothing else. Use the Venn diagram below:
Math
7
3
11
17
Band
Choir
How many students take exactly 2 classes with Mr. Green?
There are a couple of different ways to approach this problem, but one possible method is to use the formula:
n(A ∩ B) = n(A) + n(B) - n(AUB)
where n represents the number of elements in a set, and A and B are any two sets. A ∩ B represents the intersection of A and B (i.e. the elements that are in both sets), and AUB represents the union of A and B (i.e. all the elements that are in either set or both).
Using this formula, we can find the number of students who take exactly 2 classes with Mr. Green by looking at the overlap between the different sets in the Venn diagram. Specifically, we want to find n(Band ∩ Math), n(Choir ∩ Math), and n(Band ∩ Choir), since those are the three sets that represent students taking exactly 2 classes with Mr. Green.
Starting with n(Band ∩ Math), we know that there are 9 students who have Mr. Green for math and nothing else, so we can subtract that from the total number of students in the math circle:
n(Band ∩ Math) = 11 - 9 = 2
Next, we can use the formula to find n(Band ∩ Choir):
n(Band ∩ Choir) = n(Band) + n(Choir) - n(BandUC)
where BandUC represents the students who are in both Band and Choir (i.e. the overlap between those circles). Looking at the diagram, we can see that n(Band) = 25 and n(Choir) = 17, and n(BandUC) = 3, so:
n(Band ∩ Choir) = 25 + 17 - 3 = 39
Finally, we can find n(Choir ∩ Math) using a similar method:
n(Choir ∩ Math) = n(Choir) + n(Math) - n(ChoirUC)
where ChoirUC represents the overlap between Choir and Math. From the diagram, we can see that n(Math) = 11 and n(ChoirUC) = 3, so:
n(Choir ∩ Math) = 17 + 11 - 3 = 25
Therefore, the total number of students who take exactly 2 classes with Mr. Green is:
n(Band ∩ Math) + n(Band ∩ Choir) + n(Choir ∩ Math) = 2 + 39 + 25 = 66
So there are 66 students who take exactly 2 classes with Mr. Green.