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There are five seniors in a class. For each situation, write how the binomial formula is used to calculate the probability. a) In how many ways can I choose one senior to represent the group? b) In how many ways can I choose two seniors to represent the group? c) In how many ways can I choose three seniors to represent the group? d) In how many ways can I choose four seniors to represent the group? e) In how many ways can I choose five seniors to represent the group?

Daniel King

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Ross Pratt on December 1, 2018

A) The answer is 5.nCr = n! / (r! (n - r)!)n - number of things in order to be selected fromr - the number of the elect thingsThere are five seniors of a class: n = 5I choose one senior: r = 1nCr = n! / (r! (n - r)!)5C1 = 5! / (1! (5 - 1)!) = (5 * 4 * 3 * 2 * 1) / (1 * 4!) = 120 / (4 * 3 * 2 * 1) = 120 / 24 = 5b) The answer is 10.nCr = n! / (r! (n - r)!)n - number of things in order to be selected fromr - the number of the elect thingsThere are five seniors of a class: n = 5I choose two older people: r = 2nCr = n! / (r! (n - r)!)5C2 = 5! / (2! (5 - 2)!) = (5 * 4 * 3 * 2 * 1) / ((2 * 1) * 3!) = 120 / (2 * (3 * 2 * 1)) = 120 / (2 * 6) = 120 / 12 = 10c) The answer is 10.nCr = n! / (r! (n - r)!)n - number of things in order to be selected fromr - the number of the elect thingsThere are five seniors of a class: n = 5I elect three elders: r = 3nCr = n! / (r! (n - r)!)5C3 = 5! / (3! (5 - 3)!) = (5 * 4 * 3 * 2 * 1) / ((3 * 2 * 1) * 2!) = 120 / (6 * (2 * 1)) = 120 / (6 * 2) = 120 / 12 = 10d) The answer is 5.nCr = n! / (r! (n - r)!)n - number of things in order to be selected fromr - the number of the elect thingsThere are five seniors of a class: n = 5I elect four older adults: r = 4nCr = n! / (r! (n - r)!)5C4 = 5! / (4! (5 - 4)!) = (5 * 4 * 3 * 2 * 1) / ((4 * 3 * 2 * 1) * 1!) = 120 / (24 * 1) = 120 / 24 = 5e) The answer is 1.nCr = n! / (r! (n - r)!)n - number of things in order to be selected fromr - the number of the elect thingsThere are five seniors of a class: n = 5I choose five older adults: r = 5nCr = n! / (r! (n - r)!)5V5 = 5! / (5! (5 - 5)!) = (5 * 4 * 3 * 2 * 1) / ((5 * 4 * 3 * 2 * 1) * 1!) = 120 / (120 * 1) = 120 / 120 = 1


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