Respuesta :
Using the combination formula, it is found that:
- a) There are 8 students in the group.
- b) 165 teams of three can be selected from this group.
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The order in which the students are chosen is not important, which means that the combination formula is used to solve this question.
Combination formula:
is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
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Question a:
- 3 students from a set of n, thus: [tex]C_{n,3}[/tex]
- 2 students from a set of n, thus: [tex]C_{n,2}[/tex]
Twice as many ways to choose the teams of 3 than the teams of 2, thus:
[tex]C_{n,3} = 2C_{n,2}[/tex]
[tex]\frac{n!}{3!(n-3)!} = 2\frac{n!}{2!(n-2)!}[/tex]
We have to solve for n, thus:
[tex]2!n!(n-2)! = 2(3!)(n!)(n-3)![/tex]
[tex](n-2)! = 3!(n-3)![/tex]
From the factorial concept: [tex](n-2)! = (n-2)(n-3)![/tex], thus:
[tex](n-2)! = 3!(n-3)![/tex]
[tex](n-2)(n-3)! = 3!(n-3)![/tex]
[tex]n - 2 = 3![/tex]
[tex]n - 2 = 6[/tex]
[tex]n = 8[/tex]
There are 8 students in the group.
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Question b:
First we have to find the total number of students, using the same logic as item a.
[tex]C_{n,3} = 2C_{n,2}[/tex]
[tex]\frac{n!}{3!(n-3)!} = 3\frac{n!}{2!(n-2)!}[/tex]
[tex]2!n!(n-2)! = 3n!3!(n-3)![/tex]
[tex]2(n-2)(n-3)! = 18(n-3)![/tex]
[tex]n - 2 = 9[/tex]
[tex]n = 11[/tex]
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The number of teams of three is:
[tex]C_{11,3} = \frac{11!}{3!8!} = 165[/tex]
165 teams of three can be selected from this group.
A similar problem is given at https://brainly.com/question/23302762
