QuestionName: $\qquad$
8. Cora, Berlin, Shaylene, Keegan, Brooklyn, Alyssa, Hannah and Madi watch a movie together and sit in 8 adjacent seats. In how many ways can this be done under each condition? [3 marks] a. No restrictions $$8!=102,20$$ b. Cora and Berlin sit beside each other? $$\begin{array}{l} \text { rin sit beside each other? } \\ 7!=5040 \times 6! \end{array}$$ c. If Shaylene won't sit next to Brooklyn? $6!$ $-720$

Studdy Solution
To find the number of arrangements where Shaylene does not sit next to Brooklyn, we first calculate the total arrangements without restriction and then subtract the arrangements where they are sitting together.
First, calculate the arrangements where Shaylene and Brooklyn sit beside each other by treating them as a single "block":
Arrange the 7 units (Shaylene-Brooklyn block and the other 6 individuals):
$$7! = 5,040$$
Within the Shaylene-Brooklyn block, they can switch places:
$$2! = 2$$
Thus, the number of arrangements where they sit together is:
$$7! \times 2! = 5,040 \times 2 = 10,080$$
Now, subtract this from the total unrestricted arrangements:
$$8! - (7! \times 2!) = 40,320 - 10,080 = 30,240$$
The solutions for each condition are: a. $40,320$ b. $10,080$ c. $30,240$