If the Math Olympiad Club consists of 12 students, how many different teams of 3 students can be formed for competitions?

To solve this problem, we need to use the concept of combinations in mathematics. A combination is a selection of items without considering the order.

The formula for combinations is:

C(n, k) = n! / [k!(n-k)!]

where:

• n is the total number of items,
• k is the number of items to choose,
• "!" denotes factorial, which means multiplying all positive integers up to that number.

In this case, we have:

• n = 12 (the total number of students in the Math Olympiad Club), and
• k = 3 (the number of students in each team).

So, we can substitute these values into the formula:

C(12, 3) = 12! / [3!(12-3)!]

Calculating the factorials:

12! = 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 3! = 3 × 2 × 1 (12-3)! = 9! = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

Substituting these values back into the formula:

C(12, 3) = (12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) / [(3 × 2 × 1) × (9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)]

Simplifying the equation by canceling out the common factors:

C(12, 3) = (12 × 11 × 10) / (3 × 2 × 1)

C(12, 3) = 220

So, 220 different teams of 3 students can be formed from a club of 12 students.