in a running competition, a bronze, silver an...
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 13 girls are competing, how many different ways could the six medals possibly be given out?
Answer
# Explanation:
## Step1: Calculate ways for boys' medals
We use the permutation formula \(P(n,r)=\frac{n!}{(n - r)!}\), where \(n = 9\) (number of boys) and \(r=3\) (number of boys' medals). So \(P(9,3)=\frac{9!}{(9 - 3)!}=\frac{9!}{6!}=9\times8\times7 = 504\).
## Step2: Calculate ways for girls' medals
Using the permutation formula again with \(n = 13\) (number of girls) and \(r = 3\) (number of girls' medals). So \(P(13,3)=\frac{13!}{(13 - 3)!}=\frac{13!}{10!}=13\times12\times11=1716\).
## Step3: Calculate total ways
By the multiplication - principle, the total number of ways to give out the six medals is the product of the number of ways to give out the boys' medals and the number of ways to give out the girls' medals. So the total number of ways is \(504\times1716 = 865864\).
# Answer:
865864