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Mathematics 14 Online
OpenStudy (anonymous):

From 5 people, how many committees can be made with a) exactly 3 people, b) at least 3 people?

OpenStudy (anonymous):

a.5c3

OpenStudy (anonymous):

is it how many committees or in how many ways a commitee is formed?

OpenStudy (akashdeepdeb):

This is a Combinations problem because they want to know How Many COMMITTEES can be made and not in How Many WAYS can it. :D So the first one has 3 people to be SELECTED from 5 people So the first answer should be: \[^5C_3\] And for the second part There can be 3 people OR 4 people OR all the 5 people. So the number of committees would be the selection of all the three added up. So it'd be \[^5C_3~~ ~~OR~~~~ ^5C_4~~~~~ OR~~~~ ^5C_5\] \[^5C_3~~ ~~+~~~ ^5C_4~~~~~ +~~~~ ^5C_5\] Understood? :)

OpenStudy (anonymous):

@AkashdeepDeb : that's perfect... :)

OpenStudy (akashdeepdeb):

:)

OpenStudy (anonymous):

huhhh

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