Understanding the University Medal Problem

This problem involves calculating the number of students who received medals in certain combinations of sports at a university. We are given information about the total number of students who received at least one medal, the total medals given in each sport (basketball, football, and volleyball), and the number of students who received medals in all three sports. We need to find the number of students who received medals in at least two of the three sports.

Defining the Sets

Let's use sets to represent the students who received medals in each sport:

• B: The set of students who received a medal in basketball.
• F: The set of students who received a medal in football.
• V: The set of students who received a medal in volleyball.

We are given the following information based on the problem statement:

• Total students who received at least one medal (union of the sets): $|B \cup F \cup V| = 15x$
• Number of medals in basketball: $|B| = 5x$
• Number of medals in football: $|F| = 4x + 15$
• Number of medals in volleyball: $|V| = x + 25$
• Number of students who received medals in all three sports (intersection of all three sets): $|B \cap F \cap V| = x$

Applying the Principle of Inclusion-Exclusion

The principle of inclusion-exclusion for three sets helps us relate the union, individual sets, and their intersections. The formula is:

$\quad |B \cup F \cup V| = |B| + |F| + |V| - (|B \cap F| + |B \cap V| + |F \cap V|) + |B \cap F \cap V|$

We can substitute the given values into this formula:

$\quad 15x = (5x) + (4x + 15) + (x + 25) - (|B \cap F| + |B \cap V| + |F \cap V|) + x$

Calculating the Sum of Pairwise Intersections

Let's simplify the equation from the previous step:

$\quad 15x = 5x + 4x + 15 + x + 25 - (|B \cap F| + |B \cap V| + |F \cap V|) + x$

Combine the terms with x and the constant terms:

$\quad 15x = (5x + 4x + x + x) + (15 + 25) - (|B \cap F| + |B \cap V| + |F \cap V|)$

$\quad 15x = 11x + 40 - (|B \cap F| + |B \cap V| + |F \cap V|)$

Now, let's isolate the sum of the pairwise intersections:

$\quad |B \cap F| + |B \cap V| + |F \cap V| = 11x + 40 - 15x$

$\quad |B \cap F| + |B \cap V| + |F \cap V| = 40 - 4x$

This sum represents the total count of students who received medals in the overlap of any two sports, where students in all three sports are counted multiple times (once for each pair they are in). Specifically, students in exactly two sports are counted once in this sum, and students in all three sports are counted three times.

Finding Students with Medals in At Least Two Sports

The question asks for the number of students who received medals in "at least two of three sports". This includes students who received medals in exactly two sports and students who received medals in exactly three sports.

Let's break down the sum of pairwise intersections:

$\quad |B \cap F| = |\text{only } B \cap F| + |B \cap F \cap V|$

$\quad |B \cap V| = |\text{only } B \cap V| + |B \cap F \cap V|$

$\quad |F \cap V| = |\text{only } F \cap V| + |B \cap F \cap V|$

Summing these equations:

$\quad (|B \cap F| + |B \cap V| + |F \cap V|) = (|\text{only } B \cap F| + |\text{only } B \cap V| + |\text{only } F \cap V|) + 3 \times |B \cap F \cap V|$

We know $|B \cap F| + |B \cap V| + |F \cap V| = 40 - 4x$ and $|B \cap F \cap V| = x$. Substitute these values:

$\quad 40 - 4x = (|\text{only } B \cap F| + |\text{only } B \cap V| + |\text{only } F \cap V|) + 3x$

The term $(|\text{only } B \cap F| + |\text{only } B \cap V| + |\text{only } F \cap V|)$ represents the number of students who received medals in exactly two sports.

So, the number of students in exactly two sports is:

$\quad |\text{exactly two sports}| = 40 - 4x - 3x = 40 - 7x$

The number of students in exactly three sports is $|B \cap F \cap V| = x$.

The number of students in at least two sports is the sum of those in exactly two sports and those in exactly three sports:

$\quad |\text{at least two sports}| = |\text{exactly two sports}| + |\text{exactly three sports}|$

$\quad |\text{at least two sports}| = (40 - 7x) + x$

$\quad |\text{at least two sports}| = 40 - 6x$

Conclusion

The number of students who received medals in at least two of the three sports is $40 - 6x$.

Revision Table for University Sports Medals

Additional Information on Set Theory and Medals

This problem is a classic application of set theory, specifically using Venn diagrams and the principle of inclusion-exclusion to count elements in overlapping sets. Let $n(A)$ denote the number of elements in set A.

• Venn Diagrams: A Venn diagram is a visual representation of sets and their relationships. For three sets, it typically shows three overlapping circles within a rectangle (representing the universal set). Each region in the diagram corresponds to a specific combination of sets (e.g., only B, B and F but not V, all three).
• Inclusion-Exclusion Principle: This principle provides a systematic way to count the number of elements in the union of multiple sets by adding the sizes of the individual sets, subtracting the sizes of all pairwise intersections, adding back the sizes of all three-way intersections, and so on. The formula for three sets used in this problem is a direct application of this principle.
• Exactly vs. At Least: It's important to distinguish between "exactly two" and "at least two".
  • "Exactly two" refers to students who are in the intersection of two sets but not in the third set (e.g., in B and F, but not V). The sum of these is $|\text{only } B \cap F| + |\text{only } B \cap V| + |\text{only } F \cap V|$.
  • "At least two" refers to students who are in exactly two sets OR in exactly three sets. This is the sum of the "exactly two" group and the "exactly three" group ($|B \cap F \cap V|$).

Understanding these concepts is crucial for solving problems involving overlapping categories like students receiving multiple sports medals.