1. Problem: At a luncheon party for 37 people, given various counts of requests for fried rice (F), salad (S), and shredded beef (B), find: (i) the number of people requesting all three dishes; (ii) number requesting exactly two dishes; (iii) number requesting only one dish. 2. Let's denote: - $n(F)=25$, $n(S)=15$, $n(B)=20$ - $n(F \cap S)=7$, $n(F \cap B)=13$, $n(S \cap B)=8$ - $n(F \text{ only})=10$, $n(S \text{ only})=5$ 3. First, find $n(F \cap S \cap B)$ (all three). Using inclusion-exclusion for three sets: $$ n(F \cup S \cup B) = n(F) + n(S) + n(B) - n(F \cap S) - n(F \cap B) - n(S \cap B) + n(F \cap S \cap B) $$ Given all 37 requested at least one dish, so $n(F \cup S \cup B) = 37$. So, $$37 = 25 + 15 + 20 - 7 - 13 - 8 + n(F \cap S \cap B)$$ $$37 = 60 - 28 + n(F \cap S \cap B)$$ $$37 = 32 + n(F \cap S \cap B)$$ $$n(F \cap S \cap B) = 37 - 32 = 5$$ 4. Number requesting exactly two dishes is: $$n(F \cap S) + n(F \cap B) + n(S \cap B) - 3 \times n(F \cap S \cap B)$$ $$= 7 + 13 + 8 - 3 \times 5 = 28 - 15 = 13$$ 5. Number requesting only one dish is given as: $$n(F \text{ only}) + n(S \text{ only}) + n(B \text{ only})$$ We know $n(F \text{ only})=10$, $n(S \text{ only})=5$. Find $n(B \text{ only})$: $$n(B) = n(B \text{ only}) + n(F \cap B) + n(S \cap B) - n(F \cap S \cap B)$$ $$20 = n(B \text{ only}) + 13 + 8 - 5$$ $$20 = n(B \text{ only}) + 16$$ $$n(B \text{ only}) = 4$$ 6. Total only one dish: $$10 + 5 + 4 = 19$$ --- 7. Problem: College staff club beer brands. Given intersections and only counts, find: (i) number who drank Star (S); (ii) Trophy (T); (iii) Club (C); (iv) total members. 8. Data: - $n(S \cap T) = 5$, $n(S \cap C) = 7$, $n(T \cap C) = 10$ - $n(S \text{ only})=6$, $n(T \text{ only})=11$, $n(C \text{ only})=5$ - $n(S \cap T \cap C) =3$ 9. Use inclusion-exclusion to find totals: $$n(S) = n(S \text{ only}) + n(S \cap T) + n(S \cap C) - n(S \cap T \cap C)$$ $$= 6 + 5 + 7 - 3 = 15$$ $$n(T) = n(T \text{ only}) + n(S \cap T) + n(T \cap C) - n(S \cap T \cap C)$$ $$= 11 + 5 + 10 - 3 = 23$$ $$n(C) = n(C \text{ only}) + n(S \cap C) + n(T \cap C) - n(S \cap T \cap C)$$ $$= 5 + 7 + 10 - 3 = 19$$ 10. Total members: $$n(S \cup T \cup C) = n(S) + n(T) + n(C) - n(S \cap T) - n(S \cap C) - n(T \cap C) + n(S \cap T \cap C)$$ $$= 15 + 23 + 19 - 5 - 7 - 10 + 3 = 57 - 22 + 3 = 38$$ --- 11. Problem: In a school with 80 students playing hockey (H) or football (F), with conditions: - $n(F) = 5 + 2 n(H)$ - $n(H \cap F) = 15$ - All play at least one game. Find: (i) $n(F)$ (ii) number who play football but not hockey (iii) number who play hockey but not football 12. Let $h = n(H)$, $f = n(F)$. Given $f = 5 + 2h$ Total players: $$n(H \cup F) = h + f - n(H \cap F) = 80$$ Substitute $f$: $$h + (5 + 2h) - 15 = 80$$ $$3h - 10 = 80$$ $$3h = 90$$ $$h = 30$$ Then, $$f = 5 + 2 \times 30 = 65$$ 13. Number who play football but not hockey: $$n(F) - n(H \cap F) = 65 - 15 = 50$$ 14. Number who play hockey but not football: $$n(H) - n(H \cap F) = 30 - 15 = 15$$ --- 15. Problem: Sets $P = \{x : 0 \leq x \leq 3\}$ and $Q = \{x : 3 < x < 2\}$. Find $P \cup Q$. 16. Notice $Q = \{ x : 3 < x < 2 \}$ is empty (no real $x$ satisfies $3 < x < 2$). Thus: $$P \cup Q = P = \{ x : 0 \leq x \leq 3 \}$$ --- 17. Problem: Given $P=\{3,9,11,13\}$ and $Q=\{3,7,9,15\}$ subsets of universal $U=\{1,3,7,9,11,13,15\}$. Find $P \cap Q$. 18. Elements common to both: $$P \cap Q = \{3,9\}$$ --- 19. Problem: Sets $P=\{2,5,8\}$, $Q=\{2,3,5,7\}$, $U=\{1,2,3,4,5,6,7,8,9,10\}$. Find $P \cap Q$. 20. Common elements: $$P \cap Q = \{2,5\}$$ --- 21. Problem: $P=\{2x : 1 \leq x \leq 5\}$, $Q=\{x+1 : 2 \leq x \leq 8\}$ with $x$ integer, universal $U=\{x : 1 \leq x \leq 10\}$. Find: - (a) $P$ and $Q$ explicit sets - (b) $P \cup Q$ 22. Compute $P$: $x=1,2,3,4,5$ gives $$P = \{2,4,6,8,10\}$$ 23. Compute $Q$: $x=2$ to $8$ gives $$Q = \{3,4,5,6,7,8,9\}$$ 24. Union: $$P \cup Q = \{2,3,4,5,6,7,8,9,10\}$$ --- 25. Problem: 50 candidates sat exam, 31 passed chemistry (C), 29 passed physics (P), 3 failed both. Find number passed chemistry only. 26. Since 3 failed both, number passing at least one is: $$50 - 3 = 47$$ 27. Using inclusion-exclusion: $$n(C \cup P) = n(C) + n(P) - n(C \cap P) = 47$$ Solve for $n(C \cap P)$: $$31 + 29 - n(C \cap P) = 47$$ $$60 - n(C \cap P) = 47$$ $$n(C \cap P) = 13$$ 28. Chemistry only: $$n(C) - n(C \cap P) = 31 - 13 = 18$$ --- 29. Problem: Sets $P$, $Q$ with $n(P)=10$, $n(Q)=6$. Find smallest possible $n(P \cap Q)$. 30. The smallest intersection happens when $Q$ has minimal overlap with $P$: Since $Q$ size is 6 and $P$ size is 10, the smallest intersection is: $$\max(0, n(P) + n(Q) - n(U))$$ Since $n(U)$ unknown, minimum intersection is 0 if universal set is large enough. But for non-empty $P,Q$, the smallest possible $n(P \cap Q)$ is 0. Final Answers: (i) 5 (ii) 13 (iii) 19 (iv) 15, 23, 19, 38 (for problem 9) (v) Football=65, Football only=50, Hockey only=15 (vi) $P \cup Q = \{x : 0 \leq x \leq 3\}$ (vii) $P \cap Q=\{3,9\}$ (viii) $P \cap Q=\{2,5\}$ (ix) $P=\{2,4,6,8,10\}$, $Q=\{3,4,5,6,7,8,9\}$, $P \cup Q=\{2,3,4,5,6,7,8,9,10\}$ (x) Chemistry only = 18 (xi) Smallest $n(P \cap Q) = 0$