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catgirl17
 2 years ago
a four person committee is chosen at random from a group of 15 people. how many different committees are possible?
catgirl17
 2 years ago
a four person committee is chosen at random from a group of 15 people. how many different committees are possible?

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amistre64
 2 years ago
Best ResponseYou've already chosen the best response.115.14.13.12.11!  4.3.2.11! 15.14.13.12  4.3.2 15.7.13

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0is this permutation? or combination @amistre64 ?

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1combo since it doesnt suggest positions for the 4

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1its just a group of 4 people with no specific need for position abc = bca in this case

catgirl17
 2 years ago
Best ResponseYou've already chosen the best response.0where did you get the 15x7x13 ?

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.015.14.13.12  4.3.2.1 15.14.13.6  4.3. 15.14.13.2  4 15.14.13  2 15.7.13

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0do you get it @catgirl17 ?

catgirl17
 2 years ago
Best ResponseYou've already chosen the best response.0where did the 6 come from in step 2?

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0\(\Large \color{MidnightBlue}{\Rightarrow nCr = {n! \over (n  r)!*r! } }\) Got it?

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0nCr can be represented as: \(\Large \color{MidnightBlue}{ (\matrix{n \\r}) }\)

catgirl17
 2 years ago
Best ResponseYou've already chosen the best response.0no i don't get it @lgbasallote

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0hmm but you get 15.14.13.12  4.3.2.1 right?
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