Question

In a competition, a school awarded medals in different categories. 36 medals in dance, 12 medals in dramatics and 18 medals in music. If these medals went to a total of 45 persons and only 4 persons got medals in all the three categories, how many received medals in exactly two of these categories?

• A. 7
• B. 5
• C. 13
• D. 2

Hint:

For three-set medal problems, first compute $S_2=\sum$ of pairwise intersections using inclusion-exclusion. Then subtract $3\times$ the "all three" count to get the number in exactly two categories.

Answer 1

The Correct Option is C

Step 1: Notation and given data. 
Let $D$ = Dance, $R$ = Dramatics, $M$ = Music. 
$|D|=36,\ |R|=12,\ |M|=18,\ |D\cup R\cup M|=45,\ |D\cap R\cap M|=4$. 

Step 2: Inclusion-exclusion to find the sum of pairwise intersections. 
\[ |D\cup R\cup M| = |D|+|R|+|M| - \big(|D\cap R|+|D\cap M|+|R\cap M|\big) + |D\cap R\cap M|. \] Hence \[ 45 = 36+12+18 - S_2 + 4 \ \Rightarrow\ S_2 = 36+12+18+4-45 = 25, \] where $S_2=|D\cap R|+|D\cap M|+|R\cap M|$. 

Step 3: Extract those in exactly two categories. 
Every person in all three sets is counted three times inside $S_2$, while a person in exactly two sets is counted once. 
Let $E_2$ be the number who received medals in exactly two categories. With $t=|D\cap R\cap M|=4$, \[ S_2 = E_2 + 3t \ \Rightarrow\ E_2 = S_2 - 3t = 25 - 12 = 13. \] \[ \boxed{13\ \text{persons}} \]

Answer 2

Instead of computing the sum of pairwise intersections through inclusion–exclusion, this can be solved directly using the three Venn-diagram regions: exactly-one, exactly-two, and exactly-three.

Setting up region variables. Let \(p\)= number of persons in exactly one category, \(q\)= number in exactly two categories, \(r\)= number in all three. We are given \(r=4\) and total distinct people \(p+q+r=45\), so \[ p+q=41. \quad \text{(i)} \] Each medal-category count adds up people who belong to it, counting exactly-one people once, exactly-two people twice, and exactly-three people three times: \[ 36+12+18 = p+2q+3r \ \Rightarrow\ 66 = p+2q+12 \ \Rightarrow\ p+2q=54. \quad \text{(ii)} \]

Solving for \(q\). Subtracting (i) from (ii): \[ (p+2q)-(p+q)=54-41 \ \Rightarrow\ q=13. \]

1. Option 7: Would leave \(p=34\), but then \(p+2q=34+14=48\neq54\); ruled out.
2. Option 5: Would give \(p=36\), and \(p+2q=36+10=46\neq54\); ruled out.
3. Option 13: Gives \(p=28\), and \(p+2q=28+26=54\) — matches exactly.
4. Option 2: Would give \(p=39\), and \(p+2q=39+4=43\neq54\); ruled out.

Only \(q=13\) satisfies both the total-people equation and the total-medals equation simultaneously.

So the correct answer is 13 persons.