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?
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.
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.