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If the Math Olympiad Club consists of 19 students, how many different teams of 5 students can be formed for competitions?
Answer:

If the Math Olympiad Club consists of $19$ students, how many different teams of $5$ 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

19

students, how many different teams of

5

students can be formed for competitions?

\newline

Answer:

Calculate Factorial $19$ : To determine the number of different teams of $5$ students that can be formed from $19$ students, we need to calculate the combination of $19$ students taken $5$ at a time. The formula for combinations is given by: $\newline$ $C(n, k) = \frac{n!}{k! \cdot (n - k)!}$ $\newline$ where $n$ is the total number of items, $k$ is the number of items to choose, and $!$ denotes factorial.
Calculate Factorial $5$ : First, we calculate the factorial of $19$ , which is the product of all positive integers up to $19$ : $\newline$ $19! = 19 \times 18 \times 17 \times \ldots \times 1$
Calculate Factorial $14$ : Next, we calculate the factorial of $5$ , which is the product of all positive integers up to $5$ : $\newline$ $5! = 5 \times 4 \times 3 \times 2 \times 1$
Apply Combination Formula: We also need to calculate the factorial of the difference between $19$ and $5$ , which is $14$ : $\newline$ $14! = 14 \times 13 \times 12 \times \ldots \times 1$
Simplify Expression: Now we can plug these values into the combination formula: $\newline$ $C(19, 5) = \frac{19!}{(5! \cdot 14!)}$
Perform Calculation: We simplify the expression by canceling out the common terms in the numerator and the denominator. The factorials of $14$ in $19!$ and $14!$ cancel each other out: $\newline$ $C(19, 5) = \frac{19 \times 18 \times 17 \times 16 \times 15}{5 \times 4 \times 3 \times 2 \times 1}$
Perform Calculation: We simplify the expression by canceling out the common terms in the numerator and the denominator. The factorials of $14$ in $19!$ and $14!$ cancel each other out: $\newline$ $C(19, 5) = \frac{19 \times 18 \times 17 \times 16 \times 15}{5 \times 4 \times 3 \times 2 \times 1}$ Perform the calculation: $\newline$ $C(19, 5) = \frac{19 \times 18 \times 17 \times 16 \times 15}{120}$ $\newline$ $C(19, 5) = \frac{19 \times 3 \times 17 \times 16 \times 3}{1}$ $\newline$ $C(19, 5) = 969 \times 16 \times 3$ $\newline$ $C(19, 5) = 46464$

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