1. **Problem statement:** 7a) Calculate the number of ways to assign 4 different light bulbs to 4 specific lamp sockets. 7b) Calculate the probability that Joschua randomly assigns all bulbs correctly on the first try. 8) Calculate the number of ways to assign 8 people to 8 single rooms. 9) Calculate the number of ways to arrange a necklace with 12 different beads. 10) Calculate the number of ways to arrange 5 crime novels by popularity. 11) Calculate the number of ways to assign seats to 26 students in a room with 30 single desks. 12) Calculate the number of ways 6 students can choose seats from 10 available seats in order. --- 2. **Formulas and rules:** - The number of ways to assign $n$ distinct items to $n$ distinct positions is $n!$ (factorial). - The number of ways to assign $k$ items to $n$ positions (where $k \leq n$) without repetition and order matters is the permutation $P(n,k) = \frac{n!}{(n-k)!}$. - Probability of a specific arrangement when all arrangements are equally likely is $\frac{1}{\text{total number of arrangements}}$. - For arranging beads in a necklace, if the necklace is circular and rotations are considered the same, the number of distinct arrangements is $(n-1)!$. --- 3. **Step-by-step solutions:** **7a)** Number of ways to assign 4 bulbs to 4 sockets: $$4! = 4 \times 3 \times 2 \times 1 = 24$$ **7b)** Probability all bulbs are correctly assigned on first try: Total ways = 24 (from 7a) Only 1 correct arrangement. $$P = \frac{1}{24}$$ **8)** Number of ways to assign 8 people to 8 rooms: $$8! = 40320$$ **9)** Number of ways to arrange 12 different beads in a necklace: Since rotations are considered the same, number of distinct arrangements: $$(12 - 1)! = 11! = 39916800$$ **10)** Number of ways to arrange 5 crime novels by popularity: $$5! = 120$$ **11)** Number of ways to assign seats to 26 students in 30 desks: Number of permutations of 30 desks taken 26 at a time: $$P(30,26) = \frac{30!}{(30-26)!} = \frac{30!}{4!}$$ **12)** Number of ways 6 students can choose seats from 10 seats in order: $$P(10,6) = \frac{10!}{(10-6)!} = \frac{10!}{4!}$$ --- 4. **Final answers:** - 7a) 24 - 7b) $\frac{1}{24}$ - 8) 40320 - 9) 39916800 - 10) 120 - 11) $\frac{30!}{4!}$ - 12) $\frac{10!}{4!}$ These results use factorial and permutation concepts to count arrangements and probabilities.