Question

question 1 3 pts there are 8 seats in the front row of a popular english class, and 13 eager students wishing to fill them. knowing that not all of the students can find a seat in the front row, in how many ways can these seats be assigned? (enter your answer in the box below, without using comma in the number. the program will insert comma in the number where needed when you submit your answer). question 2 3 pts a coin is flipped 10 times. how many outcomes are possible in the sample space? (enter your answer in the box below, without using comma in the number. the program will insert comma in the number where needed when you submit your answer).

Explanation:

Step1: Calculate permutations for Question 1

The formula for permutations \(P(n,r)=\frac{n!}{(n - r)!}\), where \(n = 13\) (number of students) and \(r=8\) (number of seats).

$$P(13,8)=\frac{13!}{(13 - 8)!}=\frac{13!}{5!}=13\times12\times11\times10\times9\times8\times7\times6$$

$$13\times12 = 156$$

,

$$156\times11=1716$$

,

$$1716\times10 = 17160$$

,

$$17160\times9=154440$$

,

$$154440\times8 = 1235520$$

,

$$1235520\times7=8648640$$

,

$$8648640\times6=51891840$$

Step2: Calculate for Question 2

Each coin flip has 2 outcomes. For \(n = 10\) flips, use the multiplication principle. The number of outcomes in the sample space is \(2^n\).

$$2^{10}=2\times2\times2\times2\times2\times2\times2\times2\times2\times2 = 1024$$

Answer:

Question 1: \(51891840\)
Question 2: \(1024\)