Finding the intersection of 3 sets when given all information (except the intersection) 
Info:
104 students were asked if they like math, science or humanities.
35 don't like either
21 students like math only
17 students like science only
4 students like humanities only
15 students like math and humanities
13 students like math and science
17 students like humanities and science

Question:
how many students like all three subjects?

Work:
I'm not even sure what the formula is for finding n(A∩B∩C) is. But when I used the following formula, this is what I got:
formula: n(A∩B∩C) = n(A) + n(B) + n(C) - n(A∪B∪C)
established facts:
where n(A) = 49 (21 + 13 + 15)
where n(B) = 47 (13 + 17 + 17)
where n(C) = 36 (15 + 17 + 4)
where n(A∪B∪C) = 69 (104 [total] - 35 [neither])
substitution:
n(A∩B∩C) = 49 + 47 + 36 - 69
n(A∩B∩C) = 132 - 69
n(A∩B∩C) = 63?!
Could someone please let me know what I am doing wrong?? Or if there is a formula that is applicable to this particular situation (missing intersection) could someone please let me know of that??
Because the intersection cannot be 63...
 A: In general using the Inclusion-Exclusion Principle,
$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C|-|B \cap C| + |A \cap B \cap C| \implies$
$|A \cap B \cap C| = |A \cup B \cup C|  -(|A| + |B| + |C|) + |A \cap B| + |A \cap C|+|B \cap C|$.
In this case, your conditions are a bit different as we have the strict conditions of "only" for $A,B,C$ so the subtraction becomes easier for those terms. Then using a Venn Diagram one can see that $A \cap B,A \cap C,B \cap C \subseteq A \cap B \cap C$, and are oversubstracted. Hence now we know how to substract so that we have the following:
$-2|A \cap B \cap C| = (104-35)-(21+17+4)-(15+13+17)=-18 \implies$
$|A \cap B \cap C| = 9$.
What was done was this: Cancel out easily the elements (students) strictly in singleton sets, then substract the $2$-intersections so that we are left with $69 - (STUFF)= |A \cup B \cup C| - (STUFF) = -3|A \cap B \cap C|.$
A: $104-35=69$ students like at least one subject.
Then, $69-(21+17+4)=27$ students like at least two subjects.
But all of the students that like all three (call this $n$) also like at least two, so we have counted those students three times by adding $15+13+17 = 45$ (once for each of the three groups). We only want to count them once, so we subtract out $2n$:
$45 - 2n = 27 \implies n = 9.$
