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In a running competition, a bronze, silver and gold medal must be given to the top three girls and top three boys. If 4 boys and 6 girls are competing, how many different ways could the six medals possibly be given out?
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In a running competition, a bronze, silver and gold medal must be given to the top three girls and top three boys. If $4$ boys and $6$ girls are competing, how many different ways could the six medals possibly be given out? $\newline$ Answer:

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Compound events: find the number of outcomes

Full solution

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

4

boys and

6

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. There are $4$ boys competing for $3$ medals (gold, silver, and bronze). This is a permutation problem because the order in which the medals are awarded matters. $\newline$ The number of ways to award the medals to the boys is given by the permutation formula $P(n, k) = \frac{n!}{(n-k)!}$ , where $n$ is the total number of items, and $k$ is the number of items to choose. $\newline$ For the boys, $n = 4$ and $k = 3$ . $\newline$ $P(4, 3) = \frac{4!}{(4-3)!}$ $\newline$ $= \frac{4!}{1!}$ $\newline$ $= \frac{(4 \times 3 \times 2 \times 1)}{(1)}$ $\newline$ $3$ $0$
Calculate Girls' Medal Permutations: Next, we calculate the number of ways the medals can be given out to the girls. There are $6$ girls competing for $3$ medals. $\newline$ Using the permutation formula again, $P(n, k) = \frac{n!}{(n-k)!}$ , for the girls, $n = 6$ and $k = 3$ . $\newline$ $P(6, 3) = \frac{6!}{(6-3)!}$ $\newline$ $= \frac{6!}{3!}$ $\newline$ $= \frac{(6 \times 5 \times 4 \times 3 \times 2 \times 1)}{(3 \times 2 \times 1)}$ $\newline$ $= (6 \times 5 \times 4)$ $\newline$ $= \(120\)
Calculate Total Medal Distribution: Now, 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 for boys and girls are independent events, we multiply the number of ways for boys by the number of ways for girls.\(\newline\)Total number of ways \(=\) Ways for boys \(\times\) Ways for girls\(\newline\)\(= 24 \times 120\)\(\newline\)\(= 2880\)

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