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Suppose there are 24 kids in a sports competition.

(a) How many different sets of winners of gold, silver, and bronze medals are there?
(b) If three of them are to get (unranked) medals, how many different sets of winners
are there?

User Hora
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1 Answer

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Final answer:

To find the number of different ordered sets of medal winners, we use permutations, while unordered sets of winners require combinations. Using the permutation formula, there are 24 × 23 × 22 different ordered sets of winners. For the unordered sets, the combination formula is used to find the number of possible groups.

Step-by-step explanation:

The question is asking how many different sets of winners for medals are possible in a sports competition with 24 kids. This problem falls into the category of permutations and combinations within the field of mathematics.

Part A: Ordered Sets of Winners

To find the number of different ordered sets of winners (gold, silver, and bronze), we use permutations since the order matters. The formula for permutations is expressed as P(n, r) = n! / (n-r)!, where n is the total number of items to choose from, and r is the number of items we are choosing. In this case, for medals, n is 24 and r is 3. So the calculation is P(24, 3) = 24! / (24-3)! = 24! / 21! = 24 × 23 × 22.

Part B: Unordered Sets of Winners

To calculate the number of different sets of three unranked winners, we use combinations, as the order does not matter. The formula for combinations is C(n, r) = n! / (r! × (n-r)!), where again n is the total number of items and r is the number of items being chosen. Here, n is 24, and r is 3, which gives us C(24, 3) = 24! / (3! × (24-3)!) = 24! / (3! × 21!).

User Tranquilized
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