Is there a certain formula or method for solving "set" problems like these?

Question

For example...

If out of 120 students,
60 like prof. A
40 like only prof B
&
79 don't like prof B,
how many like both professors?

...besides trying to come up with a venn diagram?

Please show me how. Thank you very much.

can you show me how, please?

Answer 1

Truth Table approach
A , B , n
0 , 0 , 20 = 120 - 60 - 40
0 , 1 , 40
1 , 0 , 59 = 79 - 20
1 , 1 , 1 = 120 - 20 - 40 - 59

20 + 40 + 59 + 1 = 120

Text approach
If out of 120 students,
60 like prof. A, so
60 do not like A
40 like B and do not like A
20 do not like B and do not like A
79 don't like prof B, so only
41 like B
Only 1 who likes B can like A

You can also use a Karnaugh Map, which is essentially a "squared up" Venn diagram.

Answer 2

Well, what I usually do is create all mutually exclusive and exhaustive cases. For example, if liking is '+' and dislike is '-', our possibilities are: +A-B (like prof A, don't like prof B) , -A+B , -A-B , and +A+B. None of these have anything in common, and they cover the complete universal set. Now, as given to us,

-A+B = 40 (like only B)
+A+B + +A-B = 60 (like A) ........... (this eq is redundant in this case, but make all eqs in general.)
+A-B + -A-B = 79 (dont like B)
&
+A-B + -A+B + -A-B + +A+B = 120 (all)

solving these equations will give you the answer. ( I calculated 1)

Answer 3

Yes, the results on cardinality of sets
n(A) + n(B) = n(AUB) + n(A intersection B)

you would also need to use the DeMorgan Laws
(A intersection B)' = A' U B'
(AUB)' = A' intersection B'

btw here AUB would mean "A or B" not "A and B" A intersection B is "A and B"

and yes, it is MUCH easier to use a venn

Answer 4

60= n(A)
40= n(B and notA)= n(A or B) - n(A)
79= n(not B) = 120- n(B), or n(B) = 41

n(B and A) = n(B) + n(A) - n(A or B)
= 41 - 40
=1

Answer 5

it's still easier to use a venn, though.