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If the Math Olympiad Club consists of 16 students, how many different teams of 3 students can be formed for competitions?
Answer:

If the Math Olympiad Club consists of $16$ students, how many different teams of $3$ students can be formed for competitions? $\newline$ Answer:

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Compound events: find the number of outcomes

Full solution

Q. If the Math Olympiad Club consists of

16

students, how many different teams of

3

students can be formed for competitions?

\newline

Answer:

Calculate Combinations Formula: To determine the number of different teams of $3$ students that can be formed from $16$ students, we need to calculate the combinations of $16$ students taken $3$ at a time. This is denoted as $16C3$ , which is the number of ways to choose $3$ students from a group of $16$ without regard to order. $\newline$ The formula for combinations is: $\newline$ $nCr = \frac{n!}{r! \times (n - r)!}$ $\newline$ where $n$ is the total number of items, $r$ is the number of items to choose, and “ $16$ $0$ ” denotes factorial.
Calculate Factorials: First, we calculate the factorial of $16$ , which is $16! = 16 \times 15 \times 14 \times \ldots \times 1$ . $\newline$ Next, we calculate the factorial of $3$ , which is $3! = 3 \times 2 \times 1$ . $\newline$ Then, we calculate the factorial of $(16 - 3)$ , which is $13! = 13 \times 12 \times \ldots \times 1$ .
Plug Values into Formula: Now we can plug these values into the combinations formula: $\newline$ $16C3 = \frac{16!}{3! \times (16 - 3)!}$ $\newline$ $= \frac{16!}{3! \times 13!}$ $\newline$ $= \frac{16 \times 15 \times 14 \times 13!}{3 \times 2 \times 1 \times 13!}$ $\newline$ The $13!$ in the numerator and denominator cancel each other out.
Perform Multiplication and Division: After canceling out $13!$ , we are left with: $\newline$ $16C3 = \frac{16 \times 15 \times 14}{3 \times 2 \times 1}$ $\newline$ Now we perform the multiplication and division: $\newline$ $= \frac{16 \times 15 \times 14}{6}$ $\newline$ $= 3360 / 6$ $\newline$ $= 560$

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