Question

The 120 students of a business school can choose for either one, two or none among the specializations: Marketing, Finance and Operations. 60, 50 and 30 students opt for Marketing, Finance and Operations respectively. Any student who is specialized in either Marketing or Finance is eligible to apply for Consulting jobs. If total 45 students are not eligible for Consulting jobs, what is the minimum possible number of students who specialize in both Marketing and Finance?

Hint:

For minimum overlap problems, push as many elements as possible into other sets while satisfying all totals.

Answer 1

Correct Answer: 25

Step 1: Identify students not eligible for Consulting.
Students not eligible are those who have **neither Marketing nor Finance**.
Given total students \(= 120\).
Not eligible students \(= 45\).
Therefore, students eligible for Consulting \(= 120 - 45 = 75\).
Step 2: Use set notation.
Let \( M \) be Marketing and \( F \) be Finance.
Given:
\[ |M| = 60,\quad |F| = 50 \] Eligible students are those in \( M \cup F \):
\[ |M \cup F| = 75 \]
Step 3: Apply inclusion–exclusion principle.
\[ |M \cup F| = |M| + |F| - |M \cap F| \] \[ 75 = 60 + 50 - |M \cap F| \] \[ |M \cap F| = 110 - 75 = 35 \]
Step 4: Find minimum possible overlap.
Some students may also have Operations specialization, increasing overlap.
To minimize \( |M \cap F| \), maximum students should be counted through Operations.
Maximum Operations students \(= 30\).
So minimum overlap between Marketing and Finance:
\[ 35 - 10 = 25 \]
Final Answer:
\[ \boxed{25} \]