A committee of 9 is to be chosen from 10 seniors, 8

246,960 choices.

Pembahasan

This question involves calculating the number of ways to form a committee with a specific composition from different groups of students.

This is a problem of combinations, as the order in which students are chosen for the committee does not matter.

Steps to solve:
1.  Identify the total number of students in each group: Seniors (10), Juniors (8), Sophomores (7).
2.  Identify the required number of students from each group for the committee: Seniors (4), Juniors (3), Sophomores (2).
3.  Calculate the number of ways to choose seniors:
    This is a combination problem: C(n, k) = n! / (k!(n-k)!)
    Number of ways to choose 4 seniors from 10 = C(10, 4)
    C(10, 4) = 10! / (4!(10-4)!) = 10! / (4!6!) = (10 × 9 × 8 × 7) / (4 × 3 × 2 × 1) = 10 × 3 × 7 = 210.

4.  Calculate the number of ways to choose juniors:
    Number of ways to choose 3 juniors from 8 = C(8, 3)
    C(8, 3) = 8! / (3!(8-3)!) = 8! / (3!5!) = (8 × 7 × 6) / (3 × 2 × 1) = 8 × 7 = 56.

5.  Calculate the number of ways to choose sophomores:
    Number of ways to choose 2 sophomores from 7 = C(7, 2)
    C(7, 2) = 7! / (2!(7-2)!) = 7! / (2!5!) = (7 × 6) / (2 × 1) = 7 × 3 = 21.

6.  Calculate the total number of choices for the committee:
    Multiply the number of ways to choose from each group (using the multiplication principle).
    Total choices = (Choices for Seniors) × (Choices for Juniors) × (Choices for Sophomores)
    Total choices = 210 × 56 × 21.

7.  Perform the multiplication:
    210 × 56 = 11,760
    11,760 × 21 = 246,960

Answer:
The number of choices if the committee is to have 4 seniors, 3 juniors, and 2 sophomores is 246,960.