In a running competition, a bronze, silver and gold medal must be given to the top three girls and top three boys. If $5$ boys and $14$ girls are competing, how many different ways could the six medals possibly be given out? $\newline$ Answer:

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Q. In a running competition, a bronze, silver and gold medal must be given to the top three girls and top three boys. If

5

boys and

14

girls are competing, how many different ways could the six medals possibly be given out?

\newline

Answer:

Calculate Boys' Medal Distribution: First, we need to calculate the number of ways the medals can be given out to the boys. Since there are $5$ boys and $3$ medals (gold, silver, bronze), we use permutations because the order matters (gold is different from silver, which is different from bronze). $\newline$ The number of ways to award $3$ medals to $5$ boys is given by the permutation formula: $\newline$ $P(n, k) = \frac{n!}{(n - k)!}$ $\newline$ where $n$ is the total number of boys and $k$ is the number of medals.
Calculate Girls' Medal Distribution: For the boys, we have: $\newline$ $P(5, 3) = \frac{5!}{(5 - 3)!}$ $\newline$ $= \frac{5!}{2!}$ $\newline$ $= \frac{(5 \times 4 \times 3 \times 2 \times 1)}{(2 \times 1)}$ $\newline$ $= 5 \times 4 \times 3$ $\newline$ $= 60$ $\newline$ So, there are $60$ ways to award the medals to the boys.
Calculate Total Medal Distribution: Next, we calculate the number of ways the medals can be given out to the girls. There are $14$ girls and $3$ medals. $\newline$ Using the permutation formula again: $\newline$ $P(n, k) = \frac{n!}{(n - k)!}$ $\newline$ where $n$ is the total number of girls and $k$ is the number of medals.
Calculate Total Medal Distribution: Next, we calculate the number of ways the medals can be given out to the girls. There are $14$ girls and $3$ medals. $\newline$ Using the permutation formula again: $\newline$ $P(n, k) = \frac{n!}{(n - k)!}$ $\newline$ where $n$ is the total number of girls and $k$ is the number of medals.For the girls, we have: $\newline$ $P(14, 3) = \frac{14!}{(14 - 3)!}$ $\newline$ $= \frac{14!}{11!}$ $\newline$ $= \frac{(14 \times 13 \times 12 \times 11!)}{11!}$ $\newline$ $= 14 \times 13 \times 12$ $\newline$ $= 2184$ $\newline$ So, there are $3$ $0$ ways to award the medals to the girls.
Calculate Total Medal Distribution: Next, we calculate the number of ways the medals can be given out to the girls. There are $14$ girls and $3$ medals. $\newline$ Using the permutation formula again: $\newline$ $P(n, k) = \frac{n!}{(n - k)!}$ $\newline$ where $n$ is the total number of girls and $k$ is the number of medals.For the girls, we have: $\newline$ $P(14, 3) = \frac{14!}{(14 - 3)!}$ $\newline$ $= \frac{14!}{11!}$ $\newline$ $= \frac{(14 \times 13 \times 12 \times 11!)}{11!}$ $\newline$ $= 14 \times 13 \times 12$ $\newline$ $= 2184$ $\newline$ So, there are $3$ $0$ ways to award the medals to the girls.Finally, we need to find the total number of ways the six medals can be given out to both the boys and the girls. Since the medal distribution to the boys and girls are independent events, we multiply the number of ways for the boys by the number of ways for the girls.
Calculate Total Medal Distribution: Next, we calculate the number of ways the medals can be given out to the girls. There are $14$ girls and $3$ medals. $\newline$ Using the permutation formula again: $\newline$ $P(n, k) = \frac{n!}{(n - k)!}$ $\newline$ where $n$ is the total number of girls and $k$ is the number of medals.For the girls, we have: $\newline$ $P(14, 3) = \frac{14!}{(14 - 3)!}$ $\newline$ $= \frac{14!}{11!}$ $\newline$ $= \frac{(14 \times 13 \times 12 \times 11!)}{11!}$ $\newline$ $= 14 \times 13 \times 12$ $\newline$ $= 2184$ $\newline$ So, there are $3$ $0$ ways to award the medals to the girls.Finally, we need to find the total number of ways the six medals can be given out to both the boys and the girls. Since the medal distribution to the boys and girls are independent events, we multiply the number of ways for the boys by the number of ways for the girls.The total number of ways the six medals can be given out is: $\newline$ Total ways = Ways for boys $3$ $1$ Ways for girls $\newline$ $3$ $2$ $\newline$ $3$ $3$ $\newline$ So, there are $3$ $4$ different ways the six medals can be given out.