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There are 5 people in a raffle drawing. Two raffle winners each win gift cards. Each gift card is the same. How many ways are there to choose the winners?Decide if the situation involves a permutation or combination, and then find the number of ways to choose the winners.A.Permutation; number of ways = 20B.Combination; number of ways = 20C.Permutation; number of ways = 10D.Combination; number of ways = 10

Question

There are 5 people in a raffle drawing. Two raffle winners each win gift cards. Each gift card is the same. How many ways are there to choose the winners?Decide if the situation involves a permutation or combination, and then find the number of ways to choose the winners.A.Permutation; number of ways = 20B.Combination; number of ways = 20C.Permutation; number of ways = 10D.Combination; number of ways = 10

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Solution

This situation involves a combination, not a permutation, because the order in which the winners are chosen does not matter.

In a combination, the order of selection does not matter, whereas in a permutation, it does. Here, it doesn't matter who is chosen first or second for the gift cards, just that two people are chosen.

The formula for combinations is C(n, r) = n! / [r!(n - r)!], where n is the total number of options, r is the number of options chosen at a time, and "!" denotes factorial.

Here, n = 5 (the number of people in the raffle) and r = 2 (the number of winners).

So, the number of ways to choose the winners is C(5, 2) = 5! / [2!(5 - 2)!] = (54321) / [(21)(321)] = 10.

So, the correct answer is D. Combination; number of ways = 10.

This problem has been solved

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