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Math  /  Discrete

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!=102,208!=102,20 b. Cora and Berlin sit beside each other?  rin sit beside each other? 7!=5040×6!\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!6! −720-720

Studdy Solution

STEP 1

1. There are 8 people: Cora, Berlin, Shaylene, Keegan, Brooklyn, Alyssa, Hannah, and Madi.
2. They are to be seated in 8 adjacent seats.
3. We need to calculate the number of seating arrangements under different conditions.

STEP 2

1. Calculate the total number of unrestricted seating arrangements.
2. Calculate the number of arrangements where Cora and Berlin sit beside each other.
3. Calculate the number of arrangements where Shaylene does not sit next to Brooklyn.

STEP 3

To find the total number of unrestricted seating arrangements, we calculate the factorial of the total number of people. This is because each person can sit in any of the 8 seats, and the order matters.
8!=8×7×6×5×4×3×2×1=40,320 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40,320

STEP 4

To find the number of arrangements where Cora and Berlin sit beside each other, we treat them as a single "block" or "unit". This reduces the problem to arranging 7 units (the Cora-Berlin block and the other 6 individuals).
First, arrange the 7 units:
7!=7×6×5×4×3×2×1=5,040 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5,040
Within the Cora-Berlin block, Cora and Berlin can switch places, giving us 2 arrangements for the block:
2!=2 2! = 2
Thus, the total number of arrangements is:
7!×2!=5,040×2=10,080 7! \times 2! = 5,040 \times 2 = 10,080

STEP 5

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 7! = 5,040
Within the Shaylene-Brooklyn block, they can switch places:
2!=2 2! = 2
Thus, the number of arrangements where they sit together is:
7!×2!=5,040×2=10,080 7! \times 2! = 5,040 \times 2 = 10,080
Now, subtract this from the total unrestricted arrangements:
8!−(7!×2!)=40,320−10,080=30,240 8! - (7! \times 2!) = 40,320 - 10,080 = 30,240
The solutions for each condition are: a. 40,320 40,320 b. 10,080 10,080 c. 30,240 30,240

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