In a running competition, a bronze, silver and gold medal must be given to the top three girls and top three boys. If 11 boys and 10 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 11 boys and 10 girls are competing, how many different ways could the six medals possibly be given out?
Solution
To solve this problem, we need to calculate the number of ways the medals can be given out separately for boys and girls, and then multiply these two numbers together.
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For the boys, there are 11 competitors and 3 medals to be given out. The number of ways to give out these medals is calculated as the permutation of 11 items taken 3 at a time. This is calculated as 11P3 = 11! / (11-3)! = 11109 = 990.
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For the girls, there are 10 competitors and 3 medals to be given out. The number of ways to give out these medals is calculated as the permutation of 10 items taken 3 at a time. This is calculated as 10P3 = 10! / (10-3)! = 1098 = 720.
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To find the total number of ways the six medals can be given out, we multiply the number of ways for the boys and the girls. So, 990 * 720 = 712,800.
Therefore, there are 712,800 different ways the six medals could possibly be given out.
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