A university undergraduate has to take 5 courses from a pool of 4 maths and 3 physics courses. In how many ways can he select his course so that he takes at least two courses in physics?
To solve this problem, we can use combinations and permutations.
First, let's calculate the total number of ways the student can choose 5 courses from the 7 available courses (4 maths + 3 physics):
Total number of ways = C(7, 5) = 7! / (5! * (7-5)!) = 21
Next, we need to subtract the number of ways the student can choose only 0 or 1 physics course (since we want the student to take at least two physics courses).
Number of ways to choose 0 physics courses = C(4, 5) = 0 (since there are only 4 math courses available)
Number of ways to choose 1 physics course = C(3, 4) = 3 (choosing 1 physics course out of 3 available)
Therefore, the number of ways the student can select his courses so that he takes at least two courses in physics is:
Total number of ways - Number of ways to choose 0 or 1 physics course = 21 - 0 - 3 = 18
So, the student can select his courses in 18 different ways.
Add up how many for each possible number of physics courses.
4C2*3C2 + 4C1*3C3
= 6*3 + 4*1
= 22