In a class of 300 students in an M.Tech programme, each student is required to take at least one subject from the following three:
M600: Advanced Engineering Mathematics
C600: Computational Methods for Engineers
E600: Experimental Techniques for Engineers
The registration data for the M.Tech class shows that 100 students have taken M600, 200 students have taken C600, and 60 students have taken E600. What is the maximum possible number of students in the class who have taken all the above three subjects?

Question

Prepp Team · Accepted Answer

This problem involves finding the maximum overlap between three sets (subjects) given the total number of elements (students) and the size of each set.

Define Variables and Given Data

Let the sets represent students enrolled in the subjects:

$M$: Students enrolled in M600: Advanced Engineering Mathematics
$C$: Students enrolled in C600: Computational Methods for Engineers
$E$: Students enrolled in E600: Experimental Techniques for Engineers

We are given:

Total number of students, $N = 300$.
Number of students in M600, $|M| = 100$.
Number of students in C600, $|C| = 200$.
Number of students in E600, $|E| = 60$.
Every student takes at least one subject, so $|M \cup C \cup E| = 300$.

We need to find the maximum possible value for the number of students taking all three subjects, which is $|M \cap C \cap E|$.

Apply Inclusion-Exclusion Principle

The Principle of Inclusion-Exclusion for three sets is:

$|M \cup C \cup E| = |M| + |C| + |E| - (|M \cap C| + |M \cap E| + |C \cap E|) + |M \cap C \cap E|$

Substitute the known values:

$300 = 100 + 200 + 60 - (|M \cap C| + |M \cap E| + |C \cap E|) + |M \cap C \cap E|$ $300 = 360 - (|M \cap C| + |M \cap E| + |C \cap E|) + |M \cap C \cap E|$

Rearranging the terms to find the sum of pairwise intersections:

$|M \cap C| + |M \cap E| + |C \cap E| = 360 - 300 + |M \cap C \cap E|$ $|M \cap C| + |M \cap E| + |C \cap E| = 60 + |M \cap C \cap E|$

Maximize Triple Intersection

Let $x = |M \cap C \cap E|$.

Let $y_{MC}$ be the number of students taking only M and C (but not E).

Let $y_{ME}$ be the number of students taking only M and E (but not C).

Let $y_{CE}$ be the number of students taking only C and E (but not M).

The pairwise intersections can be expressed as:

$|M \cap C| = y_{MC} + x$
$|M \cap E| = y_{ME} + x$
$|C \cap E| = y_{CE} + x$

Substitute these into the equation derived above:

$(y_{MC} + x) + (y_{ME} + x) + (y_{CE} + x) = 60 + x$ $y_{MC} + y_{ME} + y_{CE} + 3x = 60 + x$ $y_{MC} + y_{ME} + y_{CE} + 2x = 60$

Derive Constraints and Find Maximum Value

To maximize $x$, we need to minimize the sum $y_{MC} + y_{ME} + y_{CE}$. Since the number of students must be non-negative, the minimum possible value for $y_{MC}, y_{ME}, y_{CE}$ is 0.

Therefore, the minimum value for $y_{MC} + y_{ME} + y_{CE}$ is 0.

Substituting this minimum value into the equation:

$0 + 2x = 60$ $2x = 60$ $x = 30$

Verify Feasibility

We must check if $x=30$ is possible given the constraints on the number of students in each subject.

If $x=30$ and $y_{MC} + y_{ME} + y_{CE} = 0$, it implies $y_{MC}=0, y_{ME}=0, y_{CE}=0$.

Let's check the counts for each subject:

M600 ($|M|=100$): The number of students in M is composed of those taking only M, M&C only, M&E only, and M&C&E. So, $|M| = (\text{M only}) + y_{MC} + y_{ME} + x$. $100 = (\text{M only}) + 0 + 0 + 30 \implies (\text{M only}) = 70$. This is valid ($\ge 0$).
C600 ($|C|=200$): $|C| = (\text{C only}) + y_{MC} + y_{CE} + x$. $200 = (\text{C only}) + 0 + 0 + 30 \implies (\text{C only}) = 170$. This is valid ($\ge 0$).
E600 ($|E|=60$): $|E| = (\text{E only}) + y_{ME} + y_{CE} + x$. $60 = (\text{E only}) + 0 + 0 + 30 \implies (\text{E only}) = 30$. This is valid ($\ge 0$).

The total number of students is the sum of students in each category:

$(\text{M only}) + (\text{C only}) + (\text{E only}) + y_{MC} + y_{ME} + y_{CE} + x$ $70 + 170 + 30 + 0 + 0 + 0 + 30 = 300$

This sum matches the total number of students given ($300$). Thus, the value $x=30$ is feasible.