Sets - Cardinal Numbers of Three Finite Sets-advanced

In a survey of 60 people, it was found that 25 people read newspaper $H$, 26 read newspaper $T$, and 26 read newspaper $I$. 9 read both $H$ and $I$, 11 read both $H$ and $T$, 8 read both $T$ and $I$, and 3 read all three newspapers. Find: (i) The number of people who read at least one newspaper. (ii) The number of people who read exactly one newspaper.

Given: $n(H) = 25$ $n(T) = 26$ $n(I) = 26$ $n(H \cap I) = 9$ $n(H \cap T) = 11$ $n(T \cap I) = 8$ $n(H \cap T \cap I) = 3$

(i) To find the number of people who read at least one newspaper, we calculate $n(H \cup T \cup I)$: $n(H \cup T \cup I) = n(H) + n(T) + n(I) - n(H \cap T) - n(T \cap I) - n(H \cap I) + n(H \cap T \cap I)$ $n(H \cup T \cup I) = 25 + 26 + 26 - 11 - 8 - 9 + 3$ $n(H \cup T \cup I) = 77 - 28 + 3 = 52$

(ii) To find the number of people who read exactly one newspaper: $\text{Exactly one} = n(H) + n(T) + n(I) - 2[n(H \cap T) + n(T \cap I) + n(H \cap I)] + 3n(H \cap T \cap I)$ $\text{Exactly one} = (25 + 26 + 26) - 2[11 + 8 + 9] + 3(3)$ $\text{Exactly one} = 77 - 2[28] + 9$ $\text{Exactly one} = 77 - 56 + 9 = 30$

The first part uses the standard inclusion-exclusion formula for three sets. The second part uses the derived formula for elements in exactly one set by removing elements belonging to two or more intersections.

In a class of 50 students, 20 take Mathematics, 25 take Physics, and 30 take Chemistry. 8 take Math and Physics, 10 take Physics and Chemistry, and 7 take Math and Chemistry. If every student takes at least one subject, find how many students take all three subjects.

Given: $n(U) = 50$ Since every student takes at least one subject, $n(M \cup P \cup C) = 50$. $n(M) = 20, n(P) = 25, n(C) = 30$ $n(M \cap P) = 8, n(P \cap C) = 10, n(M \cap C) = 7$

Using the formula: $n(M \cup P \cup C) = n(M) + n(P) + n(C) - n(M \cap P) - n(P \cap C) - n(M \cap C) + n(M \cap P \cap C)$ $50 = 20 + 25 + 30 - 8 - 10 - 7 + n(M \cap P \cap C)$ $50 = 75 - 25 + n(M \cap P \cap C)$ $50 = 50 + n(M \cap P \cap C)$ $n(M \cap P \cap C) = 50 - 50 = 0$

We substitute the known values into the union formula. In this specific case, the result shows that there are zero students taking all three subjects, indicating the intersections perfectly balance the total student count.