<s>[INST] How do you find the minimum percentage who have studied all four subjects? At college, 70% students studied Maths, 75% students studied English, 85% studied french and 80% studied german. What percentage at least must have studied all 4?
 [/INST] Using the Generalized Principle of Inclusion-Exclusion, we compute:
$N(0)=100$
$N(1)=70+75+80+85=310$
$N(2)=x_1+x_2+x_3+x_4+x_5+x_6=X$
$N(3)=y_1+y_2+y_3+y_4= Y$
$N(4)=Z$
$100\binom{0}{0}-310\binom{1}{0}+X\binom{2}{0}-Y\binom{3}{0}+Z\binom{4}{0}= 0 $ Study Nothing
$\hphantom{100\binom{0}{1}-}310\binom{1}{1}-X\binom{2}{1}+Y\binom{3}{1}-Z\binom{4}{1}=D$ Study One subject
$\hphantom{100\binom{0}{1}-310\binom{1}{1}+}X\binom{2}{2}-Y\binom{3}{2}+Z\binom{4}{2}=C$ Study two subjects
$\hphantom{100\binom{0}{1}-310\binom{1}{1}+X\binom{2}{2}-}Y\binom{3}{3}-Z\binom{4}{3}=B$ Study three subjects
$\hphantom{100\binom{0}{1}-310\binom{1}{1}+X\binom{2}{2}-Y\binom{3}{3}+}Z\binom{4}{4}=A$ Study four subjects. 
$x_i$ = study any two subjects
$y_i$ = study any three subjects
$100 - 310 + X - Y + Z = 0 => 100 - 310 + C + 3B + 6A - B - 4A + A = 0 =>C + 2B + 3A = 210$ ... (1)
$310 - 2X + 3Y - 4Y = D => D = 310 - 2(C+3B+6A) + 3(B+4A) - 4A = 310 - 2C - 3B - 4A$
$X - 3Y + 6Z = C => X = C +3Y - 6Z = C + 3(B+4A) - 6A = C + 3B +6A$
$Y - 4Z = B => Y = B + 4A$
$Z = A$
After working backwards, we have gotten to a point we have reduced it to three variables,namely, percentage of students who study two subjects (C), percentage of students who study three subjects (B) and finally percentage of students who study four subjects (A) with following equation $ C + 2B + 4A = 210$.  If we have C and B we could get A. But I still wonder how you could get the minimum A with just four numbers in the question.
Thanks
Satish
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