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How many 5 person committees can be selected from nine people? - Answers

To determine how many 5-person committees can be formed from nine people, you can use the combination formula ( C(n, k) = \frac{n!}{k!(n-k)!} ), where ( n ) is the total number of people, and ( k ) is the number of people to choose. In this case, ( n = 9 ) and ( k = 5 ). Thus, the number of ways to select the committee is ( C(9, 5) = \frac{9!}{5!(9-5)!} = \frac{9!}{5!4!} = 126 ). Therefore, there are 126 different 5-person committees possible.



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How many 5 person committees can be selected from nine people? - Answers

https://math.answers.com/math-and-arithmetic/How_many_5_person_committees_can_be_selected_from_nine_people

To determine how many 5-person committees can be formed from nine people, you can use the combination formula ( C(n, k) = \frac{n!}{k!(n-k)!} ), where ( n ) is the total number of people, and ( k ) is the number of people to choose. In this case, ( n = 9 ) and ( k = 5 ). Thus, the number of ways to select the committee is ( C(9, 5) = \frac{9!}{5!(9-5)!} = \frac{9!}{5!4!} = 126 ). Therefore, there are 126 different 5-person committees possible.



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https://math.answers.com/math-and-arithmetic/How_many_5_person_committees_can_be_selected_from_nine_people

How many 5 person committees can be selected from nine people? - Answers

To determine how many 5-person committees can be formed from nine people, you can use the combination formula ( C(n, k) = \frac{n!}{k!(n-k)!} ), where ( n ) is the total number of people, and ( k ) is the number of people to choose. In this case, ( n = 9 ) and ( k = 5 ). Thus, the number of ways to select the committee is ( C(9, 5) = \frac{9!}{5!(9-5)!} = \frac{9!}{5!4!} = 126 ). Therefore, there are 126 different 5-person committees possible.

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      To determine how many 5-person committees can be formed from nine people, you can use the combination formula ( C(n, k) = \frac{n!}{k!(n-k)!} ), where ( n ) is the total number of people, and ( k ) is the number of people to choose. In this case, ( n = 9 ) and ( k = 5 ). Thus, the number of ways to select the committee is ( C(9, 5) = \frac{9!}{5!(9-5)!} = \frac{9!}{5!4!} = 126 ). Therefore, there are 126 different 5-person committees possible.

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