Praxis Core Academic Skills for Educators: Mathematics (5733)Statistics and ProbabilityMedium
A student is planning their course schedule and needs to select 2 elective courses from a list of 7 available courses. If the order in which the courses are chosen does not matter, how many different combinations of 2 elective courses are possible?
- A49
- B14
- C21
- D42
Show answer & explanationAnswer & explanation
Correct answer: C. 21
Since the order does not matter, this is a combination problem. The formula for combinations is C(n, k) = n! / (k! * (n-k)!), where n is the total number of items and k is the number of items to choose. C(7, 2) = 7! / (2! * (7-2)!) = 7! / (2! * 5!) = (7*6) / (2*1) = 42 / 2 = 21.
Why the other options are wrong
- A. Incorrect. This would be 7 squared, which is not relevant here.
- B. Incorrect. This might be a result if 7 * 2 was used.
- D. Incorrect. This is the result for permutations, P(7, 2), where order matters.
Combinations
A combination is a selection of items from a larger set where the order of selection does not matter.
- Formula: C(n, k) = n! / (k! * (n-k)!), where n is the total number of items, and k is the number of items to choose.
- Used when forming groups or subsets where arrangement is irrelevant.
- Always results in a smaller number of possibilities than permutations for k > 1.
Memory trick: Commit to 'N' items, 'K'eep some, but 'K'are nothing for order.