Questions: Question 10 There are 9 seats in the front row of a popular English class, and 12 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.

Solution Steps

To solve this problem, we need to determine the number of ways to choose 9 students out of 12 and then arrange them in 9 seats. This involves two steps:

1. Choosing 9 students out of 12, which is a combination problem.
2. Arranging these 9 students in 9 seats, which is a permutation problem.

The total number of ways to assign the seats is the product of these two steps.

1. Calculate the number of ways to choose 9 students out of 12 using combinations.
2. Calculate the number of ways to arrange 9 students in 9 seats using permutations.
3. Multiply the results of the two steps to get the total number of ways to assign the seats.

To determine the number of ways to choose 9 students from 12, we use the combination formula:

\[ C(n, r) = \frac{n!}{r!(n-r)!} \]

Substituting \( n = 12 \) and \( r = 9 \):

\[ C(12, 9) = \frac{12!}{9! \cdot (12-9)!} = \frac{12!}{9! \cdot 3!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220 \]

Next, we calculate the number of ways to arrange the 9 chosen students in 9 seats, which is given by the permutation formula:

\[ P(n, r) = n! \]

Since we are arranging all 9 students, we have:

\[ P(9, 9) = 9! = 362880 \]

The total number of ways to assign the seats is the product of the combinations and permutations calculated in the previous steps:

\[ \text{Total Ways} = C(12, 9) \times P(9, 9) = 220 \times 362880 = 79833600 \]

Final Answer