The problem asks us to find the number of ways to divide a group of students into smaller groups based on given conditions. Specifically, we need to calculate the combinations for the following scenarios: a. Dividing 5 students into two groups: one with 2 students and the other with 3 students. b. Dividing 9 students into two groups: one with 4 students and the other with 5 students. c. Dividing 9 students into three groups: one with 2 students, one with 3 students, and one with 4 students.

1. Problem Description

The problem asks us to find the number of ways to divide a group of students into smaller groups based on given conditions. Specifically, we need to calculate the combinations for the following scenarios:

a. Dividing 5 students into two groups: one with 2 students and the other with 3 students.

b. Dividing 9 students into two groups: one with 4 students and the other with 5 students.

c. Dividing 9 students into three groups: one with 2 students, one with 3 students, and one with 4 students.

2. Solution Steps

a. Dividing 5 students into groups of 2 and 3:

We need to choose 2 students out of 5 for the first group. The remaining 3 students will automatically form the second group. The number of ways to choose 2 students out of 5 is given by the combination formula:

$C(n, k) = \frac{n!}{k!(n-k)!}$

where $n$ is the total number of items, $k$ is the number of items to choose, and $!$ denotes the factorial.

$C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1)(3 \times 2 \times 1)} = \frac{5 \times 4}{2 \times 1} = 10$

b. Dividing 9 students into groups of 4 and 5:

We need to choose 4 students out of 9 for the first group. The remaining 5 students will automatically form the second group. The number of ways to choose 4 students out of 9 is given by the combination formula:

$C(9, 4) = \frac{9!}{4!(9-4)!} = \frac{9!}{4!5!} = \frac{9 \times 8 \times 7 \times 6 \times 5!}{(4 \times 3 \times 2 \times 1) \times 5!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = \frac{3024}{24} = 126$

c. Dividing 9 students into groups of 2, 3, and 4:

First, we choose 2 students out of 9 for the first group. Then, we choose 3 students out of the remaining (9-2) = 7 students for the second group. Finally, the remaining 4 students form the third group.

Number of ways to choose 2 students out of 9: $C(9, 2) = \frac{9!}{2!7!} = \frac{9 \times 8}{2 \times 1} = 36$

Number of ways to choose 3 students out of the remaining 7: $C(7, 3) = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$

The remaining 4 students form the third group.

The total number of ways to divide the students into three groups is:

$C(9, 2) \times C(7, 3) = 36 \times 35 = 1260$

3. Final Answer

a. 10

b. 126

c. 1260