The question is asking how many of the 30 adults do not fall into any of the 3 categories. This type of question can be solved either algebraically, using the sets rules, or using the venn diagram. I prefer using Venn Diagrams

Date given:
Total pool of adults = 30
No. of College Graduates (C) = 15
No. of Exchange Students (E) = 10
No. who are Multilingual (M) = 8

A person who is a C can be both E & M but cannot be either E or M -> n(C&E&M) = 3
Only 4 non-C are both E&M -> n(E&M) = 4+3=7
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Formula -> n(C U E U M) = n(C) + n(E) + n(M) - (n(C INT E) + n(C INT M) + n(M INT E)) + n(C INT M INT E)
-> n(C U E U M) = 15+10+8-(3+3+7)+3 = 15+10+8-13+3 = 23

Therefore the number who are not in any category = 30-23 = 7

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We can also solve this by Venn Diagram. I have attached the Venn Diagram.

Venn Diagram Calculations
From the data collected, we can find out that the since n(C) = 15 , number of those who are only C and not anything else = 15-3 =12

Number of only M = 8-4-3 = 1

Number of only E = 10-3-4 = 3

Adding all these, we have 12+1+3+4+3 = 23

Therefore the number who are not in any category = 30-23 = 7