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Mathematics 15 Online
OpenStudy (anonymous):

@amistre64 If you are dealing from a standard deck of 52 cards, how many different 5-card hands could have at least two face cards (jacks, queens, or kings?

OpenStudy (anonymous):

that did not help me at all lol

OpenStudy (anonymous):

At the bottom :p

OpenStudy (anonymous):

nvm i got how to do it thanks anyways

OpenStudy (anonymous):

Sorry ;-;

OpenStudy (amistre64):

i wonder assume we use jacks 4C2 face cards, (52-4)C3 remaining options remove the jacks to avoid counting them again assume queens 4C2 face cards, (52-8)C3 remaining options remove the queens to avoid counting them again assume kings 4C2 face cards, (52-12)C3 remaining options

OpenStudy (amistre64):

wonder if and where the logic in my thoughts fail

OpenStudy (amistre64):

theres a flaw in it ... im sure

OpenStudy (anonymous):

I did the indirect way. So with no restrictions 52C5 What you dont want it is no face card or just 1 face card 52C5-(12C0*40C5)<--No face + (12C1*40C4)<--1 Face card)=844272

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