Q.In a group of 50 students, the number of students studying French, English, Sanskrit were found to be as follows: French = 17, English = 13, Sanskrit = 15, French and English = 09, English and Sanskrit = 4, French and Sanskrit = 5, English, French and Sanskrit = 3. Find the number of students who study

The key idea is Set Operations, specifically using the Principle of Inclusion-Exclusion for three sets.

Let $F$, $E$, and $S$ denote the sets of students studying French, English, and Sanskrit, respectively.

Given:

$|F| = 17$, $|E| = 13$, $|S| = 15$

$|F \cap E| = 9$, $|E \cap S| = 4$, $|F \cap S| = 5$

$|F \cap E \cap S| = 3$

Total students in the group = 50

(i) French only

1. Students studying French only are those in $F$ but not in $E$ and not in $S$.
2. This is calculated as $|F| - (|F \cap E| + |F \cap S|) + |F \cap E \cap S|$. …

Understanding problems involving overlapping groups of people or items is a core application of Set Theory. The key idea is to systematically account for elements that belong to one, two, or all three sets, ensuring no element is counted more than once for a specific region, or missed entirely. We use the Principle of Inclusion-Exclusion, which provides a formula to calculate the size of the union of sets, and Venn diagrams, which offer a visual way to represent these overlaps.

Let's define our sets:

• $F$: The set of students studying French.
• $E$: The set of students studying English.
• $S$: The set of students studying Sanskrit.
• $U$: The universal set of all students in the group.

We are given the following cardinalities:

• Total students, $n(U) = 50$
• $n(F) = 17$
• $n(E) = 13$
• $n(S) = 15$
• $n(F \cap E) = 9$ (French and English)
• $n(E \cap S) = 4$ (English and Sanskrit)
• $n(F \cap S) = 5$ (French and Sanskrit)
• $n(F \cap E \cap S) = 3$ (English, French and Sanskrit)

We will calculate each requested quantity step-by-step. It's often helpful to first find the sizes of the "exactly two" language groups, then the "only one" language groups, and finally the union and complement.

1. French and English but not Sanskrit ($F \cap E \text{ only}$):

   This region represents students who study French and English, but specifically not Sanskrit. We find this by taking the total number of students studying both French and English and subtracting those who study all three languages.

   $n(F \cap E \text{ only}) = n(F \cap E) - n(F \cap E \cap S)$

   $n(F \cap E \text{ only}) = 9 - 3 = 6$

2. French and Sanskrit but not English ($F \cap S \text{ only}$):

   Similarly, this is the number of students studying French and Sanskrit, excluding those who also study English.

   $n(F \cap S \text{ only}) = n(F \cap S) - n(F \cap E \cap S)$

   $n(F \cap S \text{ only}) = 5 - 3 = 2$

3. English and Sanskrit but not French ($E \cap S \text{ only}$):

   This is the number of students studying English and Sanskrit, excluding those who also study French.

   $n(E \cap S \text{ only}) = n(E \cap S) - n(F \cap E \cap S)$

   $n(E \cap S \text{ only}) = 4 - 3 = 1$

   Tip

   Visualizing these regions on a Venn diagram makes it clear. The central region $n(F \cap E \cap S)$ is subtracted from each pairwise intersection to get the "exactly two" regions.

4. French only ($F \text{ only}$):

   To find the number of students who study only French, we take the total number of students studying French and subtract all the overlaps involving French. This means subtracting those who study French and English (but not Sanskrit), French and Sanskrit (but not English), and all three languages.

   $n(F \text{ only}) = n(F) - n(F \cap E \text{ only}) - n(F \cap S \text{ only}) - n(F \cap E \cap S)$

   $n(F \text{ only}) = 17 - 6 - 2 - 3 = 17 - 11 = 6$

   ›Proof

   Alternatively, a direct formula for $n(A \text{ only})$ for three sets $A, B, C$ is:

   $n(A \text{ only}) = n(A) - n(A \cap B) - n(A \cap C) + n(A \cap B \cap C)$

   Let's verify this for $n(F \text{ only})$:

   $n(F \text{ only}) = n(F) - n(F \cap E) - n(F \cap S) + n(F \cap E \cap S)$

   $n(F \text{ only}) = 17 - 9 - 5 + 3 = 17 - 14 + 3 = 3 + 3 = 6$

   Both methods yield the same result, confirming our understanding.

5. English only ($E \text{ only}$):

   Using the same logic as above for English:

   $n(E \text{ only}) = n(E) - n(E \cap F \text{ only}) - n(E \cap S \text{ only}) - n(F \cap E \cap S)$

   $n(E \text{ only}) = 13 - 6 - 1 - 3 = 13 - 10 = 3$

   Watch out

   A common mistake is to simply subtract $n(E \cap F)$ and $n(E \cap S)$ from $n(E)$. This double-counts the students who study all three languages. The formula $n(E \text{ only}) = n(E) - n(E \cap F) - n(E \cap S) + n(F \cap E \cap S)$ correctly adjusts for this.

   $n(E \text{ only}) = 13 - 9 - 4 + 3 = 13 - 13 + 3 = 3$

6. Sanskrit only ($S \text{ only}$): …