In a running competition, a bronze, silver and gold medal must be given to the top three girls and top three boys. If 12 boys and 5 girls are competing, how many different ways could the six medals possibly be given out?
Question
In a running competition, a bronze, silver and gold medal must be given to the top three girls and top three boys. If 12 boys and 5 girls are competing, how many different ways could the six medals possibly be given out?
Solution
To solve this problem, we need to use the concept of permutations.
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For the boys, we have 12 boys and we need to find out in how many ways we can give out 3 medals. This is a permutation problem because the order in which the boys receive the medals matters (gold, silver, and bronze are different). The formula for permutations is P(n, r) = n! / (n-r)!, where n is the total number of options, r is the number of options to choose from, and "!" denotes factorial. So, for the boys, we have P(12, 3) = 12! / (12-3)! = 121110 = 1320 ways.
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Similarly, for the girls, we have 5 girls and we need to find out in how many ways we can give out 3 medals. Using the same formula, we have P(5, 3) = 5! / (5-3)! = 543 = 60 ways.
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Since these are independent events (the boys' medals don't affect the girls' medals), we multiply the two results together to get the total number of ways the six medals can be given out. So, 1320 * 60 = 79200 ways.
So, there are 79200 different ways the six medals could possibly be given out.
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