# Homework Help: Principle of Inclusion and Exclusion problem

1. Apr 1, 2012

### Dougggggg

1. The problem statement, all variables and given/known data
"A sneaky registrar reports the following information about a group of about 400 students. There are 180 taking a math class, 200 taking an English class, 160 taking a biology class, and 250 in a foreign language class. 80 are enrolled in both math and English, 90 in math and biology, 120 in math and a foreign language, 70 in English and biology, 140 in English and a foreign language, 60 in biology and a foreign language. Also, there are 25 in math, English, and a foreign language, 30 in math, English and biology, 40 in math, biology, and a foreign language, and fifteen in English, biology, and a foreign language. Finally, the sum of the number of students with a course in all four subjects, plus the number of students with a course in none of the four subjects is 100. Use Theorem 2.6 (Principle of Inclusion and Exclusion) to determine the number of students that are enrolled in all four subjects simultaneously: math, biology, English, and a foreign language."

2. Relevant equations
Theorem 2.6 (Principle of Inclusion and Exclusion).

I straight copied that from wikipedia because writing out the entire thing would just be a mess with all the notation my book uses. The one thing is in my text everything on the right hand side is subtracted to the other and they add a term for all the elements in none of those sets that are relevant so to speak. I will use my books notation below and it should be easily followed. My book doesn't define it in terms of sets.

3. The attempt at a solution
Let
N=all the students
N0=all of the students in none of the classes
N(aiaj)=all the students in class "i" and "j"
a1=in math
a2=in English
a3=in biology
a4=in a foreign language

So we have...
N(a1a2a3a4)+N0=100

So using my books notation and the substitution from above.
N0=N-N(a1)-N(a2)-N(a3)...... +N(a1a2a3a4)

Which becomes..
100-N(a1a2a3a4)=400-180-200-160-250+80+90+120+70+140+60-25-30-40-15+N(a1a2a3a4)
100=60+2N(a1a2a3a4)
N(a1a2a3a4)=20

This is a problem though because there are only 15 people taking English, biology and a foreign language... Where did I go wrong at?

2. Apr 1, 2012

### awkward

I think your computation is correct, so that means the data you have been given is inconsistent. In other words, no such set of students can exist.

3. Apr 1, 2012

### Ray Vickson

Using the notation M = number taking math, E = no. taking English, B = ..biology... and F = ... foreign languages... we have:
M=180, E=200, B=160, F=250,
ME = 80, MB = 90, MF=120, EB=70, EF=140,
MEB=30, MBF=40, MEF=25, MEB=30, MBF=40, EBF=15,
and N(all 4)+ N(none) = 100. Inclusion-exclusion gives
N(at least one) = M+E+B+F-ME-MB-MF-EB-EF+MEB+MEF+MEB+MBF+EBF-MBEF
= 400 - MBEF. Thus, for a total of N students in the group we have N(none) = N - N(at least 1) = N - 400 + MBEF, hence N(none) + MBEF = N - 400 + 2MBEF = 100, giving
N + 2MBEF = 500. If N = 400, you get MBEF = 100/2 = 50. If, for some reason, you need MBEF = 15 (as you stated) you need N = 470.

RGV

4. Apr 1, 2012

### Dougggggg

Okay, how did you make the jump from the line starting with N(at least one) to the next line where you have =400-MBEF?

5. Apr 1, 2012

### Ray Vickson

I see I left out BF = 60; that will change things to N(at least 1) = 340 - MBEF (just using the inclusion-exclusion formula). Then N(none) + MBEF = 100 is N - N(at least 1) + MBEF = N - (340 - MBEF) + MBEF = 2MBEF + N - 340 = 100, so N + 2MBEF = 440.

RGV

6. Apr 1, 2012

### Dougggggg

Okay, then you got the same thing I did. I will ask my professor about it tomorrow. Since after you subtract N=400 from both sides then divide through, it comes out to 20.