1. **Define terms:** 1.a. Permutation: An arrangement of objects in a specific order. Example: Arranging 3 books on a shelf in order. 1.b. Combination: A selection of objects without regard to order. Example: Choosing 2 fruits from an apple, banana, and orange. 1.c. Factorial: The product of all positive integers up to a number $n$, denoted $n!$. Example: $4! = 4 \times 3 \times 2 \times 1 = 24$. 1.d. Mutually exclusive: Two events that cannot happen at the same time. Example: Tossing a coin results in heads or tails, not both. 2. Chris's meal choices: 2.a. Tree diagram: (User to create physically or digitally as per instructions.) 2.b. Number of different meals: $3 \times 2 \times 3 = 18$. 2.c. Meals without Eggplant and Fritter together: - Total meals: 18 - Meals with Eggplant and Fritter: $1 \times 1 \times 1 = 1$ - So, meals possible: $18 - 1 = 17$. 2.d. Meals with at least one of Artichoke, Eggplant, Hot chocolate: - Total meals: 18 - Meals without any of these: Choose from Bruschetta (appetizer), Drumsticks (main), Fritter or Gelato (dessert except Hot chocolate) - Number without these: $1 \times 1 \times 2 = 2$ - So, meals possible: $18 - 2 = 16$. 3. Number of different tests: - Multiply options: $8 \times 3 \times 2 \times 21 = 1008$. 4. Word ABSOLUTE (8 letters): 4.a. Total arrangements: $8! = 40320$. 4.b. Arrangements with A and B together: - Treat AB as one unit: now 7 units. - Arrangements: $7! = 5040$. - AB can be AB or BA: multiply by 2. - Total: $5040 \times 2 = 10080$. 4.c. Arrangements starting with vowel (A, E, O, U) and ending with A or T: - Vowels in ABSOLUTE: A, O, U, E - Ending letters: A or T - Case 1: Start A, end A (impossible, only one A) - Case 2: Start A, end T - Case 3: Start E, end A or T - Case 4: Start O, end A or T - Case 5: Start U, end A or T Calculate each case: - Fix start and end, arrange remaining 6 letters. - Number of arrangements for each case: $6! = 720$ Count valid cases: - Start A, end T: 1 case, 720 arrangements - Start E, end A: 1 case, 720 arrangements - Start E, end T: 1 case, 720 arrangements - Start O, end A: 1 case, 720 arrangements - Start O, end T: 1 case, 720 arrangements - Start U, end A: 1 case, 720 arrangements - Start U, end T: 1 case, 720 arrangements Total: $7 \times 720 = 5040$. 5. Digits {1,2,3,4,5} forming 3-digit numbers: 5.a. No digit repeated: - Choices: $5 \times 4 \times 3 = 60$. 5.b. Number must be even, no digit repeated: - Even digits: 2,4 - Last digit choices: 2 options - First digit choices: 4 (excluding last digit) - Middle digit choices: 3 (excluding first and last) - Total: $4 \times 3 \times 2 = 24$. 5.c. Number must contain at least 1 repeated digit: - Total 3-digit numbers with digits 1-5: $5^3 = 125$ - Numbers with no repeated digits: 60 (from 5.a) - So, numbers with at least one repeated digit: $125 - 60 = 65$. 6. Deck of cards with values 5-10, 4 suits: 6.a. Number of 5-card hands: - Total cards: $6 \times 4 = 24$ - Number of 5-card hands: $\binom{24}{5} = 42504$. 6.b. Hands with at least 4 red cards (Hearts and Diamonds): - Red cards: $6 \times 2 = 12$ - Cases: - 4 red + 1 black: $\binom{12}{4} \times \binom{12}{1} = 495 \times 12 = 5940$ - 5 red: $\binom{12}{5} = 792$ - Total: $5940 + 792 = 6732$. 6.c. Hands with at least one club and at least one heart: - Total hands: 42504 - Hands with no clubs: $\binom{18}{5} = 8568$ - Hands with no hearts: $\binom{18}{5} = 8568$ - Hands with no clubs and no hearts (only spades and diamonds): $\binom{12}{5} = 792$ - By inclusion-exclusion: - Hands with at least one club and one heart = $42504 - 8568 - 8568 + 792 = 25760$. 6.d. Hands with at least one spade and at least two 10s: - Number of 10s: 4 (one per suit) - At least two 10s means 2, 3, or 4 tens. - Calculate for each case and sum: Case 1: Exactly 2 tens: - Choose 2 tens: $\binom{4}{2} = 6$ - Choose 3 other cards from 20 non-10 cards: $\binom{20}{3} = 1140$ - Total: $6 \times 1140 = 6840$ Case 2: Exactly 3 tens: - Choose 3 tens: $\binom{4}{3} = 4$ - Choose 2 other cards from 20: $\binom{20}{2} = 190$ - Total: $4 \times 190 = 760$ Case 3: Exactly 4 tens: - Choose all 4 tens: 1 way - Choose 1 other card from 20: $\binom{20}{1} = 20$ - Total: $1 \times 20 = 20$ Sum: $6840 + 760 + 20 = 7620$ Now, must have at least one spade: - Count hands with at least 2 tens but no spade: - Tens without spade 10: choose 2,3,4 tens from 3 tens (non-spade tens) No spade tens: - 2 tens: $\binom{3}{2} = 3$, choose 3 from 20 - 3 tens: $\binom{3}{3} = 1$, choose 2 from 20 - 4 tens: impossible (only 3 tens without spade) Calculate no spade tens hands: - 2 tens no spade: $3 \times 1140 = 3420$ - 3 tens no spade: $1 \times 190 = 190$ - 4 tens no spade: 0 Total no spade tens: $3420 + 190 = 3610$ Hands with at least 2 tens and at least one spade: - $7620 - 3610 = 4010$. 7. Row 8 of Pascal's triangle: - Row 0 is [1] - Row 8 has 9 elements: $\binom{8}{0}, \binom{8}{1}, ..., \binom{8}{8}$ - Values: 1, 8, 28, 56, 70, 56, 28, 8, 1 - Found using $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ or by adding adjacent numbers from row 7. 8. Row 11 of Pascal's triangle: - Row 11 has 12 elements: $\binom{11}{0}, \binom{11}{1}, ..., \binom{11}{11}$ - Values: 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1 - Found similarly by combination formula or adding adjacent numbers from row 10. 9. Paths on Cartesian grid moving right and up: 9.a. From A(0,0) to B(6,3): - Total moves: $6 + 3 = 9$ - Choose positions of 3 ups (or 6 rights): $\binom{9}{3} = 84$. 9.b. From A(0,0) to C(9,10) passing through D(5,4) and E(7,8): - Paths A to D: $\binom{5+4}{4} = \binom{9}{4} = 126$ - Paths D to E: $\binom{7-5 + 8-4}{7-5} = \binom{6}{2} = 15$ - Paths E to C: $\binom{9-7 + 10-8}{9-7} = \binom{4}{2} = 6$ - Total paths: $126 \times 15 \times 6 = 11340$. 10. Survey of 200 students: Given: - Crosswords (C): 68 - Baking (B): 97 - Acting (A): 71 - A and C: 20 - A and B: 29 - B and C: 23 - None: 28 Total students enjoying at least one activity: $200 - 28 = 172$ 10.a. Number enjoying all three (A \cap B \cap C): - Use inclusion-exclusion: $$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$$ - Substitute: $$172 = 71 + 97 + 68 - 29 - 20 - 23 + |A \cap B \cap C|$$ - Calculate: $$172 = 236 - 72 + |A \cap B \cap C|$$ $$172 = 164 + |A \cap B \cap C|$$ - So, $$|A \cap B \cap C| = 172 - 164 = 8$$ 10.b. Number enjoying only acting: - Only A = $|A| - |A \cap B| - |A \cap C| + |A \cap B \cap C|$ - Substitute: $$71 - 29 - 20 + 8 = 30$$ 10.c. Number enjoying only baking: - Only B = $|B| - |A \cap B| - |B \cap C| + |A \cap B \cap C|$ - Substitute: $$97 - 29 - 23 + 8 = 53$$ 10.d. Venn diagram description: - Three overlapping circles labeled Acting, Baking, Crosswords. - Center intersection (all three): 8 - Pairwise intersections minus center: - Acting & Baking only: $29 - 8 = 21$ - Acting & Crosswords only: $20 - 8 = 12$ - Baking & Crosswords only: $23 - 8 = 15$ - Only one activity: - Acting only: 30 - Baking only: 53 - Crosswords only: $68 - 12 - 15 - 8 = 33$ - Outside all circles: 28 This completes all questions.