A college awarded 38 medals in Football, 15 in Basketball and 20 in Cricket. If these medals went to a total of 58 men and only 3 men got medals in all the 3 sports, how many received medals in exactly two of the 3 sports?

Question

Prepp Team · Accepted Answer

Analyzing the College Sports Medals Problem

The problem provides information about the number of medals awarded in three different sports: Football, Basketball, and Cricket, and the total number of unique individuals who received these medals. We are also told how many individuals received medals in all three sports. Our goal is to find the number of individuals who received medals in exactly two of the three sports.

Let's denote the sets of men who received medals in each sport:

F: Set of men who received medals in Football.
B: Set of men who received medals in Basketball.
C: Set of men who received medals in Cricket.

We are given the following information:

Number of medals in Football, $|\text{F}| = 38$.
Number of medals in Basketball, $|\text{B}| = 15$.
Number of medals in Cricket, $|\text{C}| = 20$.
Total number of distinct men who received medals (in at least one sport), $|\text{F} \cup \text{B} \cup \text{C}| = 58$.
Number of men who received medals in all three sports, $|\text{F} \cap \text{B} \cap \text{C}| = 3$.

We want to find the number of men who received medals in exactly two sports. This corresponds to the sum of the number of men in (F ∩ B but not C), (F ∩ C but not B), and (B ∩ C but not F).

Applying the Principle of Inclusion-Exclusion

The Principle of Inclusion-Exclusion for three sets is given by:

$|\text{F} \cup \text{B} \cup \text{C}| = |\text{F}| + |\text{B}| + |\text{C}| - (|\text{F} \cap \text{B}| + |\text{F} \cap \text{C}| + |\text{B} \cap \text{C}|) + |\text{F} \cap \text{B} \cap \text{C}|$

We can plug in the values we know:

$58 = 38 + 15 + 20 - (|\text{F} \cap \text{B}| + |\text{F} \cap \text{C}| + |\text{B} \cap \text{C}|) + 3$

Simplify the equation:

$58 = 73 - (|\text{F} \cap \text{B}| + |\text{F} \cap \text{C}| + |\text{B} \cap \text{C}|) + 3$

$58 = 76 - (|\text{F} \cap \text{B}| + |\text{F} \cap \text{C}| + |\text{B} \cap \text{C}|)$

Now, solve for the sum of the pairwise intersections:

$|\text{F} \cap \text{B}| + |\text{F} \cap \text{C}| + |\text{B} \cap \text{C}| = 76 - 58$

$|\text{F} \cap \text{B}| + |\text{F} \cap \text{C}| + |\text{B} \cap \text{C}| = 18$

The sum of the numbers of men in the intersections of each pair of sports is 18.

Calculating Men with Medals in Exactly Two Sports

Let:

$\text{x}$ be the number of men who received medals in exactly Football and Basketball.
$\text{y}$ be the number of men who received medals in exactly Football and Cricket.
$\text{z}$ be the number of men who received medals in exactly Basketball and Cricket.
$\text{w}$ be the number of men who received medals in all three sports. We know $\text{w} = |\text{F} \cap \text{B} \cap \text{C}| = 3$.

The intersection of any two sets includes those who received medals in all three sports. So:

$|\text{F} \cap \text{B}| = \text{x} + \text{w}$
$|\text{F} \cap \text{C}| = \text{y} + \text{w}$
$|\text{B} \cap \text{C}| = \text{z} + \text{w}$

The sum of the pairwise intersections is therefore:

$|\text{F} \cap \text{B}| + |\text{F} \cap \text{C}| + |\text{B} \cap \text{C}| = (\text{x} + \text{w}) + (\text{y} + \text{w}) + (\text{z} + \text{w}) = \text{x} + \text{y} + \text{z} + 3\text{w}$

We found that $|\text{F} \cap \text{B}| + |\text{F} \cap \text{C}| + |\text{B} \cap \text{C}| = 18$. Substituting this and the value of $\text{w}$:

$18 = \text{x} + \text{y} + \text{z} + 3(3)$

$18 = \text{x} + \text{y} + \text{z} + 9$

Now, solve for $\text{x} + \text{y} + \text{z}$, which represents the number of men who received medals in exactly two sports:

$\text{x} + \text{y} + \text{z} = 18 - 9$

$\text{x} + \text{y} + \text{z} = 9$

So, 9 men received medals in exactly two of the three sports.

Summary of Results

Let's check the numbers with a slightly different perspective. The total number of individuals who received medals is 58.

Each individual falls into one of four categories regarding these three sports:

Received medals in exactly one sport.
Received medals in exactly two sports.
Received medals in exactly three sports.
Received no medals in these sports (this category is not relevant as we are dealing with the 58 men who received *at least one* medal).

The sum of the number of medals awarded is $38 + 15 + 20 = 73$.

This sum counts individuals multiple times if they received medals in more than one sport.

Men who received medals in exactly one sport are counted once.
Men who received medals in exactly two sports are counted twice.
Men who received medals in exactly three sports are counted three times.

Let:

$N_1$ be the number of men who received medals in exactly one sport.
$N_2$ be the number of men who received medals in exactly two sports (this is what we need to find).
$N_3$ be the number of men who received medals in exactly three sports ($N_3 = 3$).

The total number of distinct men is $N_1 + N_2 + N_3 = 58$.

The sum of the individual sport medal counts is $N_1 \times 1 + N_2 \times 2 + N_3 \times 3$.

So, $73 = N_1 + 2N_2 + 3N_3$.

We have two equations:

$N_1 + N_2 + N_3 = 58$
$N_1 + 2N_2 + 3N_3 = 73$

Substitute $N_3 = 3$ into both equations:

$N_1 + N_2 + 3 = 58 \Rightarrow N_1 + N_2 = 55$
$N_1 + 2N_2 + 3(3) = 73 \Rightarrow N_1 + 2N_2 + 9 = 73 \Rightarrow N_1 + 2N_2 = 64$

Now we have a system of two linear equations with two variables:

$N_1 + N_2 = 55$
$N_1 + 2N_2 = 64$

Subtract equation (1) from equation (2):

$(N_1 + 2N_2) - (N_1 + N_2) = 64 - 55$

$N_2 = 9$

This confirms that the number of men who received medals in exactly two sports is 9.

Conclusion

Using the Principle of Inclusion-Exclusion or by setting up equations based on the number of times each person is counted, we find that 9 men received medals in exactly two of the three sports.

Category	Number of Men
Received medals in all 3 sports ($N_3$)	3
Received medals in exactly two sports ($N_2$)	9
Received medals in exactly one sport ($N_1$)	$58 - (N_2 + N_3) = 58 - (9 + 3) = 58 - 12 = 46$
Total distinct men with medals	$N_1 + N_2 + N_3 = 46 + 9 + 3 = 58$

Check sum of medals: $1 \times N_1 + 2 \times N_2 + 3 \times N_3 = 1 \times 46 + 2 \times 9 + 3 \times 3 = 46 + 18 + 9 = 73$. This matches the sum of given medals ($38 + 15 + 20 = 73$).

The number of men who received medals in exactly two sports is 9.