Combinations Question (Please provide an explanation!)

Question

1. There are 140 students in the entire 12th Grade class.
2. 52 signed up for biology.
3. 71 signed up for chemistry.
4. 40 signed up for physics.
5. 15 signed up for biology and chemistry
6. 8 signed up for chemistry and physics.
7. 11 signed up for biology and physics.
8. 2 signed up for all 3 sciences.

a) How many students are NOT taking any science?

b) Illustrate this using a Venn diagram.

Please provide an explanation, I don't understand.

Answer 1

Let

B be the set of students signed up for biology,

C be the set of students signed up for chemistry,

P be the set of students signed up for physics.

Using the notation

|S| to denote the size of a set S,

we are given:

|B| = 52,

|C| = 71,

|P| = 40,

|B∩C| = 15,

|C∩P| = 8,

|B∩P| = 11,

|B∩C∩P| = 2

If we knew how many students signed up for at least one science,

we could subtract that from 140 to get the number of students not taking

any science.

The number of students signed up for at least one science is expressed by

|B ∪ C ∪ P|

Let me just tell you the way to calculate that and only then explain it

|B ∪ C ∪ P| = |B| + |C| + |P| - |B∩C| - |C∩P| - |B∩P| + |B∩C∩P|

= 52 + 71 + 40 - 15 - 8 - 11 + 2 = 131

So 9 students are not taking any science.

For the explanation you need to draw a Venn diagram, but first draw one for B and C only.

If we wanted to calculate |B ∪ C|, we could (naively) just add |B| + |C|.

That would be wrong because that would double-count those students signed up

for both subject. So we simply subtract that number:

|B ∪ C| = |B| + |C| - |B∩C|

Now draw the Venn diagram for all three subjects.

Again we start by approximating |B ∪ C ∪ P| as |B| + |C| + |P|.

That double counts students signed for two subjects, so we improve

the approximation by

|B| + |C| + |P| - |B∩C| - |C∩P| - |B∩P|

This expression counts exactly once students taking only one subject or only two subjects.

But consider a student taking all three.

That student appears once in each of |B|, |C|, |P|, |B∩C|, |C∩P|, |B∩P|;

after adding him in three times we subtracted him three times.

Therefore we need to add in |B∩C∩P|.