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Your class is holding elections to choose a leadership committee of 4 students. If there are 30 students in the class, how many different leadership committees would it be possible to elect?

810,000

27,405

120

657,720

Respuesta :

irspow
Since order is not important we are only looking for unique combinations of 3, so we need to use the "n choose k" formula.

n!/(k!(n-k)!), n=number of elements to choose from, k is the number of choices made....in this case:

30!/(4!(30-4)!)

30!/(4!*26!)

27405

Using the combination formula, it is found that it would be possible to elect 27,405 different leadership committees.

The order in which the students are selected to the committee is not important, which means that the combination formula is used to solve this question.

Combination formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

In this problem, 4 students are chosen from a set of 30, thus:

[tex]C_{30,4} = \frac{30!}{4!26!} = 27405[/tex]

It would be possible to elect 27,405 different leadership committees.

A similar problem is given at https://brainly.com/question/24437717

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