Given n(S) = 17, n(SS) = 15, n(BDT) = 16, n(BDT ∩ SS) = 4, n(S ∩ SS) = 6, n(BDT ∩ S) = 3, n(S ∩ SS ∩ BDT) = 6, and total students = 67, determine if these values are consistent for a Venn diagram.

Let $S$ represent Science, $SS$ represent Social Studies, and $BDT$ represent B.D.T.
The total number of students is 67.
The number of students learning each subject or combination of subjects is:

• $n(S) = 17$
• $n(SS) = 15$
• $n(BDT) = 16$
• $n(BDT \cap SS) = 4$
• $n(S \cap SS) = 6$
• $n(BDT \cap S) = 3$
• $n(S \cap SS \cap BDT) = 6$

In a standard Venn diagram, the number of students learning all three subjects, $n(S \cap SS \cap BDT)$, must be less than or equal to the number of students learning any two subjects, e.g., $n(BDT \cap SS)$.
Given $n(S \cap SS \cap BDT) = 6$:

• For $n(BDT \cap SS) = 4$: The number of students learning BDT and Social Studies only (not Science) would be $n(BDT \cap SS) - n(S \cap SS \cap BDT) = 4 - 6 = -2$. This is impossible, as the number of students cannot be negative.
• For $n(BDT \cap S) = 3$: The number of students learning BDT and Science only (not Social Studies) would be $n(BDT \cap S) - n(S \cap SS \cap BDT) = 3 - 6 = -3$. This is also impossible.
• For $n(S \cap SS) = 6$: The number of students learning Science and Social Studies only (not BDT) would be $n(S \cap SS) - n(S \cap SS \cap BDT) = 6 - 6 = 0$. This is possible.

Since two of the calculated "only two subjects" regions result in negative numbers, the problem statement is contradictory under the standard interpretation.

Sometimes, problems phrase $n(A \cap B)$ to mean "only $A$ and $B$ (not $C$)". Let's test this interpretation.
Assume the given two-subject intersection numbers are for only those two subjects, and $n(S \cap SS \cap BDT) = 6$.

• $n(BDT \cap SS \text{ only}) = 4$
• $n(S \cap SS \text{ only}) = 6$
• $n(BDT \cap S \text{ only}) = 3$
• $n(S \cap SS \cap BDT) = 6$

Now, calculate the number of students learning only one subject:

• For Science only:

This is possible.

• For Social Studies only:

This is impossible, as the number of students cannot be negative. The sum of students in the intersections involving Social Studies ($6+4+6=16$) is already greater than the total number of students learning Social Studies (15).

Both the standard and common alternative interpretations of the given data lead to impossible negative numbers of students in certain categories. Therefore, the problem statement contains contradictory information and cannot be solved as written.