If the Math Olympiad Club consists of $16$ students, how many different teams of $4$ students can be formed for competitions? $\newline$ Answer:

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Q. If the Math Olympiad Club consists of

16

students, how many different teams of

4

students can be formed for competitions?

\newline

Answer:

Calculate Combination Formula: To determine the number of different teams of $4$ students that can be formed from $16$ students, we need to calculate the combination of $16$ students taken $4$ at a time. This is denoted as $16C4$ , which is the number of ways to choose $4$ students from a group of $16$ without regard to the order of selection.
Apply Combination Formula: The formula for combinations is given by: $\newline$ $nC_k = \frac{n!}{k! \cdot (n - k)!}$ $\newline$ where $n$ is the total number of items, $k$ is the number of items to choose, and “ $!$ ” denotes factorial.
Simplify Factorials: Using the formula, we calculate $16C4$ as follows: $\newline$ $16C4 = \frac{16!}{(4! * (16 - 4)!)}$ $\newline$ $16C4 = \frac{16!}{(4! * 12!)}$
Perform Arithmetic: We can simplify the factorials by canceling out the common terms in the numerator and the denominator: $\newline$ $\frac{16!}{(4! \times 12!)} = \frac{(16 \times 15 \times 14 \times 13 \times 12!)}{(4 \times 3 \times 2 \times 1 \times 12!)}$ $\newline$ The $12!$ in the numerator and denominator cancel each other out.
Final Result: After canceling, we are left with: $\newline$ egin{equation} $\newline$ \frac{ $16$ \times $15$ \times $14$ \times $13$ }{ $4$ \times $3$ \times $2$ \times $1$ } $\newline$ \end{equation}
Final Result: After canceling, we are left with: $\newline$ $(16 \times 15 \times 14 \times 13) / (4 \times 3 \times 2 \times 1)$ Now we perform the arithmetic: $\newline$ $(16 \times 15 \times 14 \times 13) / (4 \times 3 \times 2 \times 1) = (16/4) \times (15/3) \times (14/2) \times 13$ $\newline$ $(16/4) \times (15/3) \times (14/2) \times 13 = 4 \times 5 \times 7 \times 13$
Final Result: After canceling, we are left with: $\newline$ $(16 \times 15 \times 14 \times 13) / (4 \times 3 \times 2 \times 1)$ Now we perform the arithmetic: $\newline$ $(16 \times 15 \times 14 \times 13) / (4 \times 3 \times 2 \times 1) = (16/4) \times (15/3) \times (14/2) \times 13$ $\newline$ $(16/4) \times (15/3) \times (14/2) \times 13 = 4 \times 5 \times 7 \times 13$ Multiplying these numbers together gives us the total number of different teams: $\newline$ $4 \times 5 \times 7 \times 13 = 20 \times 7 \times 13$ $\newline$ $20 \times 7 \times 13 = 140 \times 13$ $\newline$ $140 \times 13 = 1820$