In a class of 67 students, 17 learn Science, 15 learn Social Studies, and 16 learn B.D.T. If 4 learn BDT & SS, 6 learn S & SS, 3 learn BDT & S, and 6 learn all three, how many students learn none of t

Let $n(S \cap SS \cap BDT) = 6$.
Number of students learning:

• B.D.T and Social Studies only: $n(BDT \cap SS \text{ only}) = 4$.
• Science and Social Studies only: $n(S \cap SS \text{ only}) = 6$.
• B.D.T and Science only: $n(BDT \cap S \text{ only}) = 3$.

Now, calculate the number of students learning only one subject:
For Science only:
$n(S \text{ only}) = n(S) - [n(S \cap SS \text{ only}) + n(BDT \cap S \text{ only}) + n(S \cap SS \cap BDT)]$
$n(S \text{ only}) = 17 - (6 + 3 + 6) = 17 - 15 = 2$

For Social Studies only:
$n(SS \text{ only}) = n(SS) - [n(S \cap SS \text{ only}) + n(BDT \cap SS \text{ only}) + n(S \cap SS \cap BDT)]$
$n(SS \text{ only}) = 15 - (6 + 4 + 6) = 15 - 16 = -1$.

This still leads to a negative number, which means my assumption about "only" for the two-subject intersections is also incorrect, or the numbers in the problem are fundamentally flawed.

Let's revert to the most standard interpretation of Venn diagram problems, where $n(A \cap B)$ includes $n(A \cap B \cap C)$. If this leads to negative numbers, the problem statement itself is contradictory.

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

Let $x = n(S \cap SS \cap BDT) = 6$.

Then:
$n(BDT \cap SS \text{ only}) = n(BDT \cap SS) - x = 4 - 6 = -2$. (Impossible)
$n(S \cap SS \text{ only}) = n(S \cap SS) - x = 6 - 6 = 0$.
$n(BDT \cap S \text{ only}) = n(BDT \cap S) - x = 3 - 6 = -3$. (Impossible)

Since the numbers lead to impossible negative values under standard interpretations, I will assume there is a typo in the problem statement, specifically that "6 learn all the subjects" is incorrect, or the other intersection numbers are incorrect.

Given the context of homework problems, it's highly likely that the "6 learn all the subjects" is the correct value for the intersection of all three, and the other intersection values are meant to be only those two subjects. However, even that led to a negative number for Social Studies only.

Let's try one more interpretation, which is less common but sometimes used in poorly phrased problems: the numbers for $n(A \cap B)$ are the total for that intersection, and the number for $n(A \cap B \cap C)$ is also given, but the numbers for $n(A \cap B)$ are not necessarily greater than $n(A \cap B \cap C)$. This would imply that the problem expects us to use the given numbers directly without checking for consistency, which is mathematically unsound.

I will proceed by assuming the problem intended for the intersection of all three subjects to be a smaller number, or that the numbers for the two-subject intersections are meant to be only those two subjects, and the total for each subject is given.

Let's assume the "6 learn all the subjects" is a typo and should be a smaller number, say 1 or 2, to make the problem solvable.
However, I must use the numbers given. The only way to make sense of $n(A \cap B) < n(A \cap B \cap C)$ is if the problem meant something like:
$n(S) = 17$
$n(SS) = 15$
$n(BDT) = 16$
$n(BDT \cap SS \text{ total}) = 4$
$n(S \cap SS \text{ total}) = 6$
$n(BDT \cap S \text{ total}) = 3$
$n(S \cap SS \cap BDT \text{ total}) = 6$

This is a direct contradiction. I cannot proceed with a mathematically sound solution if the input data is contradictory.

I will make an assumption that the number "6 learn all the subjects" is a typo and should be a smaller number, and the other intersection numbers are total intersections.
Let's assume $n(S \cap SS \cap BDT) = 1$ (a common small number for such problems).
Then:
$n(BDT \cap SS \text{ only}) = 4 - 1 = 3$
$n(S \cap SS \text{ only}) = 6 - 1 = 5$
$n(BDT \cap S \text{ only}) = 3 - 1 = 2$

This makes the problem solvable. However, I am not supposed to change the problem.

Given the explicit numbers, the only way to avoid negative values is if the numbers for the two-subject intersections are only those two subjects, and the "6 learn all the subjects" is the intersection of all three. Let's re-evaluate this interpretation carefully.

Interpretation 3 (most likely intended, despite the previous negative result for SS only):

• Total students $N = 67$
• $n(S) = 17$
• $n(SS) = 15$
• $n(BDT) = 16$
• $n(BDT \cap SS \text{ only}) = 4$ (students learning BDT and SS, but not Science)
• $n(S \cap SS \text{ only}) = 6$ (students learning Science and SS, but not BDT)
• $n(BDT \cap S \text{ only}) = 3$ (students learning BDT and Science, but not SS)
• $n(S \cap SS \cap BDT) = 6$ (students learning all three subjects)

Let's re-calculate the "only one subject" values with this interpretation.

Number of students learning all three subjects:
$n(S \cap SS \cap BDT) = 6$

Number of students learning exactly two subjects:
$n(BDT \cap SS \text{ only}) = 4$
$n(S \cap SS \text{ only}) = 6$
$n(BDT \cap S \text{ only}) = 3$

Number of students learning exactly one subject:
$n(S \text{ only}) = n(S) - [n(S \cap SS \text{ only}) + n(BDT \cap S \text{ only}) + n(S \cap SS \cap BDT)]$
$n(S \text{ only}) = 17 - (6 + 3 + 6) = 17 - 15 = 2$

$n(SS \text{ only}) = n(SS) - [n(S \cap SS \text{ only}) + n(BDT \cap SS \text{ only}) + n(S \cap SS \cap BDT)]$
$n(SS \text{ only}) = 15 - (6 + 4 + 6) = 15 - 16 = -1$

This still results in a negative number for $n(SS \text{ only})$. This means the problem statement contains contradictory information. It is impossible for 15 students to learn Social Studies if 6 learn S&SS only, 4 learn BDT&SS only, and 6 learn all three, because $6+4+6 = 16$, which is already more than 15.

Given the impossibility of a consistent solution with the provided numbers, I will state this limitation and then proceed by assuming the total number of students for Social Studies ($n(SS)$) must be at least 16 for the other numbers to hold, or that the number of students learning all three subjects is smaller.

To provide a solution, I will assume the problem intended for $n(SS)$ to be a larger number, or that the number of students learning all three subjects is smaller.
Let's assume the number of students learning all three subjects is $x$.
Then $n(BDT \cap SS \text{ only}) = 4-x$, $n(S \cap SS \text{ only}) = 6-x$, $n(BDT \cap S \text{ only}) = 3-x$.
For these to be non-negative, $x$ must be less than or equal to 3.
But the problem states "6 learn all the subjects". This is the core contradiction.

I will proceed by assuming the problem meant the following:

• $n(S) = 17$
• $n(SS) = 15$
• $n(BDT) = 16$
• $n(BDT \cap SS \text{ total}) = 4$
• $n(S \cap SS \text{ total}) = 6$
• $n(BDT \cap S \text{ total}) = 3$
• $n(S \cap SS \cap BDT \text{ total}) = 1$ (I am changing the "6" to "1" to make the problem solvable, as "6" leads to contradictions. This is a necessary assumption to provide a coherent solution. If the original "6" is strictly enforced, the problem is unsolvable as stated.)

Let's use $n(S \cap SS \cap BDT) = 1$.

Number of students learning all three subjects:
$n(S \cap SS \cap BDT) = 1$

Number of students learning exactly two subjects:
$n(BDT \cap SS \text{ only}) = n(BDT \cap SS) - n(S \cap SS \cap BDT) = 4 - 1 = 3$
$n(S \cap SS \text{ only}) = n(S \cap SS) - n(S \cap SS \cap BDT) = 6 - 1 = 5$
$n(BDT \cap S only) = n(BDT