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

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amistre64Best ResponseYou've already chosen the best response.1
15.14.13.12.11!  4.3.2.11! 15.14.13.12  4.3.2 15.7.13
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
is this permutation? or combination @amistre64 ?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
combo since it doesnt suggest positions for the 4
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
its just a group of 4 people with no specific need for position abc = bca in this case
 one year ago

catgirl17Best ResponseYou've already chosen the best response.0
where did you get the 15x7x13 ?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
15.14.13.12  4.3.2.1 15.14.13.6  4.3. 15.14.13.2  4 15.14.13  2 15.7.13
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
do you get it @catgirl17 ?
 one year ago

catgirl17Best ResponseYou've already chosen the best response.0
where did the 6 come from in step 2?
 one year ago

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

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

catgirl17Best ResponseYou've already chosen the best response.0
no i don't get it @lgbasallote
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
hmm but you get 15.14.13.12  4.3.2.1 right?
 one year ago
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