A college offers 5 different courses. If a student can choose any 2 courses, how

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Question: A college offers 5 different courses. If a student can choose any 2 courses, how many different combinations can they make?

Options:

Correct Answer: 10

Solution:

Number of combinations = 5C2 = 5! / (2!(5-2)!) = 10.

A college offers 5 different courses. If a student can choose any 2 courses, how many different combinations can they make?

A college offers 5 different courses. If a student can choose any 2 courses, how many different combinations can they make?

Step 1: Understand that the student needs to choose 2 courses from a total of 5 courses.
Step 2: Recognize that the order in which the courses are chosen does not matter. This means we are looking for combinations, not permutations.
Step 3: Use the combination formula, which is written as nCr, where n is the total number of items (courses) and r is the number of items to choose (courses). In this case, n = 5 and r = 2.
Step 4: The combination formula is nCr = n! / (r!(n-r)!). Here, we will substitute n = 5 and r = 2 into the formula.
Step 5: Calculate 5C2 using the formula: 5C2 = 5! / (2!(5-2)!) = 5! / (2! * 3!).
Step 6: Calculate the factorials: 5! = 5 × 4 × 3 × 2 × 1 = 120, 2! = 2 × 1 = 2, and 3! = 3 × 2 × 1 = 6.
Step 7: Substitute the factorial values back into the equation: 5C2 = 120 / (2 * 6).
Step 8: Calculate the denominator: 2 * 6 = 12.
Step 9: Now divide the numerator by the denominator: 120 / 12 = 10.
Step 10: Conclude that there are 10 different combinations of courses that the student can choose.

Combinations – The concept of combinations involves selecting items from a larger set where the order of selection does not matter.
Factorial – Understanding factorial notation (n!) is crucial for calculating combinations, as it represents the product of all positive integers up to n.