Explanation
$$\textbf{Step – 1: Formulating the information to use Inclusion-Exclusion Rule.}$$
$$\text{Therefore the number of men who won in all the three sports will be}$$ $$n(F \cap B \cap C)$$
$$\text{and the number of people who got medals in two of the three sports is}$$
$$n(F \cap B) + n(B \cap C) + n(C \cap F)$$
$$\text{and the total number of men is represented as}$$ $$n(F \cup B \cup C)$$.
$$\text{From the given data we have:}$$
$$n(F) = 38$$
$$n(B) = 15$$
$$n(C) = 20$$
$$n(F \cup B \cup C) = 58$$
$$n(F \cap B \cap C) = 3$$
$$\textbf{Step – 2: Apply the Inclusion-Exclusion Rule}$$.
$$\text{Using the Inclusion-Exclusion Rule we get:}$$
$$n(F \cup B \cup C) = n(F) + n(B) + n(C) - n(F \cap B) - n(B \cap C) - n(C \cap F) + n(F \cap B \cap C)$$
$$\Rightarrow 58 = 38 + 15 + 20 - n(F \cap B) - n(B \cap C) - n(C \cap F) + 3 $$
$$\Rightarrow n(F \cap B) + n(B \cap C) + n(C \cap F) = 38 + 15 + 20 + 3 - 58 $$
$$\Rightarrow n(F \cap B) + n(B \cap C) + n(C \cap F) = 18 $$
$$\therefore\text{ the number of students who have received medals in exactly two of the three sports is}$$
$${= n(F \cap B) + n(B \cap C) + n(C \cap F) - 3 \times n(F \cap B \cap C) }$$
$${= 18 - 3 \times 3 }$$
$${= 18 - 9}$$
$${= 9}$$
$$\textbf{Hence, the number of students who have received medals in exactly two of the three sports is 9}$$
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