Among 100 students, 32 study mathematics, 20 sewing physics, 45 study histology, 15 study mathematics and biology, 7 study mathematics and physics, 10 study physios and biology and 30 do net study ant of the three subjects.

a) Find the number of students staying in all three subjects.

b) Find the number of students studying exactly one of the three subjects.

Solution:

Let $A, B, C$ denote the set of students studying mathematics, physics and biology respectively.

$ \begin{aligned} |X| &=100 \\\\ |A| &=32 \\\\ |D| &=20 \\\\ |C| &=45 \\\\ |A \cap C| &=15 \\\\ |A \cap B| &=7 \\\\ |B \cap C| &=10 \\\\ \left|A^{\prime} \cap B^{\prime} \cap C^{\prime}\right| &=30 \\\\ \left|A^{\prime} \cap B^{\prime} \cap C^{\prime}\right| &=100-\mid A \cup B \cup C \\\\ \text { Dr }|A \cup B \cup C| &=100-30 \\\\ &=70 \\ \end{aligned} $

a) number of students staying in all three subjects.

$ \begin{aligned} |A \cup B \cup C| &=|A|+|B|+|C|-|| A \cap B|+| A \cap C|+| B \cap C||+|A \cap B \cap C| \\\\ 70 &=32+20+45-[7+15+10|+| A \cap B \cap C \\\\ 70 &=97-|32|+|A \cap B \cap C| \\\\ 70-65 &=|A \cap B \cap C| \\\\ \Rightarrow|A \cap B \cap C| &=5 \\ \end{aligned} $

5 students study all 3 subjects.

b) Number of students studying exactly one subject.

The number of students studying only mathematics is,

$ \begin{aligned} A|-| A \cap B|-| A \cap C|+| A \cap B \cap C \mid \\\\ &=32-7-15+5 \\\\ &=15 \\ \end{aligned} $

Number of students studying only physics is,

$ \begin{aligned} |B|-|B \cap A|-|B \cap C|+|A \cap B \cap C| \\\\ &=20-7-10+5 \\\\ &=8 \\ \end{aligned} $

The number of students studying only biology is,

$ \begin{aligned} |C|-|A \cap C|-|B \cap C|+|A \cap B \cap C| \\\\ &=45-15-10+5 \\\\ &=25 \\ \end{aligned} $

$\therefore$ Number of students studying exactly one subject,

$ =15+8+25=48 $