1. **Problem A:** Given universal set $U = \{x: 0 < x \leq 12, x \text{ is an integer}\}$ and subsets: - $A = \{x: 0 < x \leq 6\}$ - $B = \{x: 0 < x \leq 12, x \text{ is even}\}$ - $C = \{x: x \text{ is a multiple of } 3\}$ **i) List elements of A, B, and C:** - $A = \{1, 2, 3, 4, 5, 6\}$ - $B = \{2, 4, 6, 8, 10, 12\}$ (even numbers from 1 to 12) - $C = \{3, 6, 9, 12\}$ (multiples of 3 up to 12) 2. **ii) Construct the Venn Diagram:** - The Venn diagram would have three circles labeled A, B, and C inside the universal set U. - Elements are placed in intersections according to membership: - $A \cap B = \{2,4,6\}$ - $A \cap C = \{3,6\}$ - $B \cap C = \{6,12\}$ - $A \cap B \cap C = \{6\}$ - Other elements placed accordingly. 3. **iii) Find elements of $[(A \cup B) \cap C]$:** - First find $A \cup B = \{1,2,3,4,5,6,8,10,12\}$ - Then intersect with $C = \{3,6,9,12\}$ - So, $[(A \cup B) \cap C] = \{3,6,12\}$ --- 4. **Problem B:** Given 600 adults with sports participation: - Cricket (C): 200 - Tennis (T): 250 - Hockey (H): 300 - Cricket & Tennis: 80 - Tennis & Hockey: 90 - Cricket & Hockey: 70 - All three: 40 5. **Using Inclusion-Exclusion Principle:** - Number playing only one sport: $$|C \text{ only}| = |C| - |C \cap T| - |C \cap H| + |C \cap T \cap H| = 200 - 80 - 70 + 40 = 90$$ $$|T \text{ only}| = 250 - 80 - 90 + 40 = 120$$ $$|H \text{ only}| = 300 - 70 - 90 + 40 = 180$$ - Total only one sport: $$90 + 120 + 180 = 390$$ 6. **Number playing exactly two sports:** - Exactly two sports means those in two sets but not all three: $$|C \cap T| - |C \cap T \cap H| = 80 - 40 = 40$$ $$|T \cap H| - |C \cap T \cap H| = 90 - 40 = 50$$ $$|C \cap H| - |C \cap T \cap H| = 70 - 40 = 30$$ - Total exactly two sports: $$40 + 50 + 30 = 120$$ --- 7. **Problem C:** - **Permutation principle:** The number of ways to arrange $n$ distinct objects in order is $n!$. - For selecting and arranging $r$ objects from $n$, permutations are: $$P(n,r) = \frac{n!}{(n-r)!}$$ - **Combination principle:** The number of ways to select $r$ objects from $n$ without regard to order is: $$C(n,r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}$$ - **Inverse of a permutation:** Given a permutation as a bijection from a set to itself, its inverse permutation reverses the mapping, undoing the original permutation.