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Mathematics 19 Online
OpenStudy (immanuelv):

if you draw a 5-card hand from a standard deck of playing cards, how many hands could contain exactly 4 cards from the same suit?

OpenStudy (anonymous):

wait a little confused. does this mean what is the probability you get exactly 4 cards of the same suit?

OpenStudy (anonymous):

watchmath do you understand this question?

OpenStudy (anonymous):

perhaps the answer is \[\dbinom{13}{4}\times \dbinom{39}{1}\times 4\]?

OpenStudy (anonymous):

i don't really know how to intemperate "if you draw a 5-card hand"..."how many hands"

OpenStudy (immanuelv):

that is what i'm trying to figure out!

OpenStudy (anonymous):

well maybe it means how many possible hand have 4 of one suit and two of another. if so, lets start with clubs. you need 4 clubs and 1 non club. the number of ways to get 4 clubs is \[\dbinom{13}{4}\] and the number of ways you can choose one non-club is \[\dbinom{39}{1}\]

OpenStudy (anonymous):

then we do the same for each other suit hence multiply by 4

OpenStudy (immanuelv):

yes.....

OpenStudy (immanuelv):

my theory is 4(13 C4* 39C1)

OpenStudy (anonymous):

that is what i wrote so it looks good to me

OpenStudy (anonymous):

4*13*11*5*39

OpenStudy (immanuelv):

Thank you very much !!

OpenStudy (anonymous):

welcome

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