1. **Problem 3a:** Make a truth table for the statement $ (P \lor Q) \to (P \land R) $. 2. **Truth Table Setup:** We list all possible truth values for $P$, $Q$, and $R$, then compute $P \lor Q$, $P \land R$, and finally $ (P \lor Q) \to (P \land R) $. | $P$ | $Q$ | $R$ | $P \lor Q$ | $P \land R$ | $ (P \lor Q) \to (P \land R) $ | |---|---|---|---|---|---| | T | T | T | T | T | T | | T | T | F | T | F | F | | T | F | T | T | T | T | | T | F | F | T | F | F | | F | T | T | T | F | F | | F | T | F | T | F | F | | F | F | T | F | F | T | | F | F | F | F | F | T | 3. **Problem 3b:** Simplify logical statements so negation appears only before variables. (1) $\neg(P \to \neg Q)$ Recall $P \to Q \equiv \neg P \lor Q$, so: $$\neg(P \to \neg Q) = \neg(\neg P \lor \neg Q) = P \land Q$$ (2) $(\neg P \lor \neg Q) \to \neg(\neg Q \land R)$ Rewrite implication: $$(\neg P \lor \neg Q) \to (Q \lor \neg R) = \neg(\neg P \lor \neg Q) \lor (Q \lor \neg R)$$ Simplify: $$ (P \land Q) \lor Q \lor \neg R = Q \lor \neg R$$ (3) $\neg(P \to \neg Q) \lor \neg(R \land \neg R)$ From (1), $\neg(P \to \neg Q) = P \land Q$. Note $R \land \neg R$ is always false, so $\neg(R \land \neg R)$ is always true. Thus: $$ (P \land Q) \lor \text{True} = \text{True}$$ (4) "It is false that if Samuel is not a man then Christian is a woman, and that Christian is not a woman." Let $S$ = "Samuel is a man", $C$ = "Christian is a woman". The statement is: $$ \neg( (\neg S \to C) \land \neg C ) $$ Rewrite $\neg S \to C$ as $S \lor C$. So: $$ \neg( (S \lor C) \land \neg C ) = \neg( S \land \neg C \lor C \land \neg C ) = \neg( S \land \neg C \lor \text{False} ) = \neg( S \land \neg C ) = \neg S \lor C $$ 4. **Problem 3e:** Use truth tables to show if $ (P \land (P \to Q)) \to Q $ is a tautology or contradiction. | $P$ | $Q$ | $P \to Q$ | $P \land (P \to Q)$ | $ (P \land (P \to Q)) \to Q $ | |---|---|---|---|---| | T | T | T | T | T | | T | F | F | F | T | | F | T | T | F | T | | F | F | T | F | T | All values are true, so it is a tautology. 5. **Problem 4a:** Define terms. (1) Permutation: Arrangement of objects in order. (2) Combination: Selection of objects without order. (3) Factorial: Product of all positive integers up to a number $n$, denoted $n!$. (4) Combinatorics: Study of counting, arrangement, and combination of objects. 6. **Problem 4b:** (1) Ways to arrange 6 books: $$6! = 720$$ (2) Ways to select 3 students from 20: $$\binom{20}{3} = \frac{20!}{3! \times 17!} = 1140$$ (3) Ways to arrange letters in "HELLO": Letters: H(1), E(1), L(2), O(1) Total letters = 5 Number of arrangements: $$\frac{5!}{2!} = \frac{120}{2} = 60$$ 7. **Problem 5a:** License plates with 2 letters + 4 digits (0-9). (1) Second letter is 'O' or 'Q' (2 choices), last digit is 3 or 8 (2 choices). First letter: 26 choices Digits 2,3,4: each 10 choices Total plates: $$26 \times 2 \times 10 \times 10 \times 10 \times 2 = 104000$$ (2) First letter 'K' (1 choice), second letter 'A' (1 choice), last digit 5 (1 choice). Digits 3,4: each 10 choices Total plates: $$1 \times 1 \times 10 \times 10 \times 1 = 100$$ 8. **Problem 5b:** (1) Distinguishable arrangements of "MISSISSIPPI": Letters count: M(1), I(4), S(4), P(2) Total letters = 11 Number of arrangements: $$\frac{11!}{1! \times 4! \times 4! \times 2!} = 34650$$ (2) Committee of 7 tourists from Nairobi(6), London(7), New York(8) with at least 2 from each city. Possible distributions: - 2 Nairobi, 2 London, 3 New York - 2 Nairobi, 3 London, 2 New York - 3 Nairobi, 2 London, 2 New York Calculate each: $$\binom{6}{2} \times \binom{7}{2} \times \binom{8}{3} = 15 \times 21 \times 56 = 17640$$ $$\binom{6}{2} \times \binom{7}{3} \times \binom{8}{2} = 15 \times 35 \times 28 = 14700$$ $$\binom{6}{3} \times \binom{7}{2} \times \binom{8}{2} = 20 \times 21 \times 28 = 11760$$ Total ways: $$17640 + 14700 + 11760 = 44100$$