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

View step-by-step help

Home

Math Problems

Grade 8

Experimental probability

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

14

boys and

5

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

\newline

Answer:

Calculate Boys' Medal Combinations: First, we need to calculate the number of ways the medals can be given out to the boys. Since there are $14$ 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 $14$ 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 Combinations: For the boys, we calculate: $\newline$ $P(14, 3) = \frac{14!}{(14 - 3)!}$ $\newline$ $= \frac{14!}{11!}$ $\newline$ $= 14 \times 13 \times 12$ $\newline$ $= 2184$ $\newline$ There are $2,184$ 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 $5$ 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 Ways: For the girls, we calculate: $\newline$ $P(5, 3) = \frac{5!}{(5 - 3)!}$ $\newline$ $= \frac{5!}{2!}$ $\newline$ $= 5 \times 4 \times 3$ $\newline$ $= 60$ $\newline$ There are $60$ different ways to award the medals to the girls.
Calculate Total Ways: For the girls, we calculate: $\newline$ $P(5, 3) = \frac{5!}{(5 - 3)!}$ $\newline$ $= \frac{5!}{2!}$ $\newline$ $= 5 \times 4 \times 3$ $\newline$ $= 60$ $\newline$ There are $60$ 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 for the boys by the number of ways for the girls. $\newline$ Total ways = Ways for boys $\times$ Ways for girls $\newline$ $= 2184 \times 60$
Calculate Total Ways: For the girls, we calculate: $\newline$ $P(5, 3) = \frac{5!}{(5 - 3)!}$ $\newline$ $= \frac{5!}{2!}$ $\newline$ $= 5 \times 4 \times 3$ $\newline$ $= 60$ $\newline$ There are $60$ 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 for the boys by the number of ways for the girls. $\newline$ Total ways = Ways for boys $\times$ Ways for girls $\newline$ $= 2184 \times 60$ Calculating the total number of ways: $\newline$ Total ways $= 2184 \times 60$ $\newline$ $= 131040$ $\newline$ So, there are $131,040$ different ways the six medals can be given out.