Question

A student is making out her schedule for next year. Her list consists of 9 required courses and 12 elective courses. She must take at least 4 courses but cannot take more than 6 courses. In how many ways can she select courses such that she takes at least one elective course?

Answer 1

[AI-generated answer — Explain Bot]

To solve this problem, we can use the concept of combinations.

First, let's consider the number of ways she can select exactly 4 courses (1 elective course and 3 required courses).

To select 1 elective course out of 12, we can use the combination formula: C(12, 1) = 12.

To select 3 required courses out of 9, we can use the combination formula again: C(9, 3) = 84.

So the total number of ways she can select exactly 4 courses is 12 * 84 = 1008.

Next, let's consider the number of ways she can select exactly 5 courses (2 elective courses and 3 required courses).

To select 2 elective courses out of 12, we can use the combination formula: C(12, 2) = 66.

To select 3 required courses out of 9, we can use the combination formula again: C(9, 3) = 84.

So the total number of ways she can select exactly 5 courses is 66 * 84 = 5544.

Finally, let's consider the number of ways she can select exactly 6 courses (3 elective courses and 3 required courses).

To select 3 elective courses out of 12, we can use the combination formula: C(12, 3) = 220.

To select 3 required courses out of 9, we can use the combination formula again: C(9, 3) = 84.

So the total number of ways she can select exactly 6 courses is 220 * 84 = 18480.

Therefore, the total number of ways she can select courses such that she takes at least one elective course is 1008 + 5544 + 18480 = 25032.

Answer 2

[AI-generated answer — Step-by-Step Bot]

To find the number of ways the student can select courses, we need to consider the different scenarios based on the number of courses taken.

Case 1: The student takes 4 courses
- In this case, the student selects 4 courses out of the 21 (9 required + 12 elective) courses.
- The number of ways to select 4 courses out of 21 is given by the combination formula: C(21, 4) = 21! / (4! * (21-4)!) = 5985

Case 2: The student takes 5 courses
- In this case, the student selects 5 courses out of the 21 courses.
- The number of ways to select 5 courses out of 21 is given by the combination formula: C(21, 5) = 21! / (5! * (21-5)!) = 53130

Case 3: The student takes 6 courses
- In this case, the student selects 6 courses out of the 21 courses.
- The number of ways to select 6 courses out of 21 is given by the combination formula: C(21, 6) = 21! / (6! * (21-6)!) = 54264

To find the total number of ways the student can select courses, we sum up the possibilities from the three cases:

Total number of ways = 5985 + 53130 + 54264 = 113,379

Therefore, the student can select courses in 113,379 different ways such that she takes at least one elective course.