Sets - Application of Sets-advanced

In a survey of $600$ students in a school, $150$ students were found to be taking tea and $225$ taking coffee, $100$ were taking both tea and coffee. Find how many students were taking neither tea nor coffee.

Let $U$ be the set of surveyed students, $T$ be the set of students taking tea, and $C$ be the set of students taking coffee. Given: $n(U) = 600$ $n(T) = 150$ $n(C) = 225$ $n(T \cap C) = 100$ First, find the number of students taking at least one drink: $n(T \cup C) = n(T) + n(C) - n(T \cap C)$ $n(T \cup C) = 150 + 225 - 100$ $n(T \cup C) = 275$ Number of students taking neither tea nor coffee: $n(T \cup C)' = n(U) - n(T \cup C)$ $\begin{array}{r} 600 \\ -275 \\ \hline 325 \end{array}$ There are $325$ students taking neither.

We use the addition theorem for two sets to find the total number of students who drink at least one beverage, then subtract this from the total number of students surveyed to find those who drink neither.

In a group of students, $25$ play cricket, $20$ play football, and $15$ play hockey. $10$ play both cricket and football, $8$ play football and hockey, and $7$ play cricket and hockey. $5$ students play all three games. Find the total number of students who play at least one game.

Let $C, F,$ and $H$ represent the sets of students playing cricket, football, and hockey respectively. Given: $n(C) = 25$ $n(F) = 20$ $n(H) = 15$ $n(C \cap F) = 10$ $n(F \cap H) = 8$ $n(C \cap H) = 7$ $n(C \cap F \cap H) = 5$ Using the formula for three sets: $n(C \cup F \cup H) = n(C) + n(F) + n(H) - [n(C \cap F) + n(F \cap H) + n(C \cap H)] + n(C \cap F \cap H)$ $n(C \cup F \cup H) = 25 + 20 + 15 - (10 + 8 + 7) + 5$ $n(C \cup F \cup H) = 60 - 25 + 5$ $n(C \cup F \cup H) = 40$ $40$ students play at least one game.

The principle of inclusion-exclusion for three sets is applied. We add individual counts, subtract double intersections to avoid double-counting, and add back the triple intersection which was subtracted one too many times.

In a class of $35$ students, $17$ have taken Mathematics, $10$ have taken Mathematics but not Economics. If every student has taken at least one subject, find the number of students who have taken Economics and the number of students who have taken Economics but not Mathematics.

Let $M$ be the set of students who took Mathematics and $E$ be the set of students who took Economics. Given: $n(M \cup E) = 35$ (since every student takes at least one) $n(M) = 17$ $n(M - E) = 10$ We know $n(M - E) = n(M) - n(M \cap E)$ $10 = 17 - n(M \cap E)$ $n(M \cap E) = 17 - 10 = 7$ Now, use the union formula to find $n(E)$: $n(M \cup E) = n(M) + n(E) - n(M \cap E)$ $35 = 17 + n(E) - 7$ $35 = 10 + n(E)$ $n(E) = 25$ Number of students who took Economics but not Mathematics: $n(E - M) = n(E) - n(M \cap E)$ $n(E - M) = 25 - 7 = 18$ So, $25$ students took Economics and $18$ took Economics only.

We use the relationship between the difference of sets and the intersection to find $n(M \cap E)$, then use the union formula for two sets to solve for the unknown set $E$.