In a running competition, a bronze, silver and gold medal must be given to the top three girls and top three boys. If $9$ 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

9

boys and

14

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

\newline

Answer:

Calculate Boys' Medal Permutations: First, we need to calculate the number of ways the medals can be given out to the boys. Since there are $9$ boys and $3$ medals (gold, silver, and bronze), we use permutations because the order in which the medals are awarded matters. $\newline$ The number of ways to award $3$ medals to $9$ 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 Permutations: For the boys, we calculate the permutation: $\newline$ $P(9, 3) = \frac{9!}{(9 - 3)!}$ $\newline$ $P(9, 3) = \frac{9!}{6!}$ $\newline$ $P(9, 3) = \frac{(9 \times 8 \times 7 \times 6!)}{6!}$ $\newline$ The factorials of $6$ cancel out, leaving us with: $\newline$ $P(9, 3) = 9 \times 8 \times 7$ $\newline$ $P(9, 3) = 504$ $\newline$ So, there are $504$ different ways to award the medals to the boys.
Calculate Total Medal Combinations: 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: $\newline$ $P(14, 3) = \frac{14!}{(14 - 3)!}$ $\newline$ $P(14, 3) = \frac{14!}{11!}$ $\newline$ $P(14, 3) = \frac{14 \times 13 \times 12 \times 11!}{11!}$ $\newline$ The factorials of $11$ cancel out, leaving us with: $\newline$ $P(14, 3) = 14 \times 13 \times 12$ $\newline$ $P(14, 3) = 2184$ $\newline$ So, there are $2184$ different ways to award the medals to the girls.
Calculate Total Medal Combinations: 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: $\newline$ $P(14, 3) = \frac{14!}{(14 - 3)!}$ $\newline$ $P(14, 3) = \frac{14!}{11!}$ $\newline$ $P(14, 3) = \frac{(14 \times 13 \times 12 \times 11!)}{11!}$ $\newline$ The factorials of $11$ cancel out, leaving us with: $\newline$ $P(14, 3) = 14 \times 13 \times 12$ $\newline$ $P(14, 3) = 2184$ $\newline$ So, there are $2184$ different ways to award the medals to the girls.Finally, to find the total number of different ways the six medals can be given out, we multiply the number of ways the boys' medals can be awarded by the number of ways the girls' medals can be awarded. $\newline$ Total ways = $P(\text{boys}) \times P(\text{girls})$ $\newline$ Total ways = $3$ $0$ $\newline$ Total ways = $3$ $1$ $\newline$ So, there are $3$ $2$ different ways the six medals can be given out.