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Mathematics 16 Online
OpenStudy (anonymous):

From a deck of 52 cards, how many 5-card hands can be formed with at least 3 diamonds?

OpenStudy (anonymous):

there are 13 diamonds divided by 3 gives 4 possible decks with at least 3 diamonds

OpenStudy (anonymous):

wut.

OpenStudy (zarkon):

A lot. What have you tried?

OpenStudy (anonymous):

im not sure, i feel like it starts off with \[? nCr 5 \]?

OpenStudy (zarkon):

do you know how many hands if there are exactly 3 diamonds?

OpenStudy (anonymous):

:/ no...

OpenStudy (anonymous):

5 hand?

OpenStudy (zarkon):

how many ways can you choose 3 cards from 13?

OpenStudy (anonymous):

so 13 Cr 3?

OpenStudy (anonymous):

o wait

OpenStudy (anonymous):

i meant 13 nCr 3

OpenStudy (zarkon):

drop the r \[\Large_{13}C_3\]

OpenStudy (zarkon):

ok how many cards are not diamonds?

OpenStudy (anonymous):

i don't know

OpenStudy (zarkon):

a card is a diamond or it is not. you have 52 total cards and 13 of them are diamonds

OpenStudy (anonymous):

39

OpenStudy (zarkon):

ok

OpenStudy (zarkon):

how many ways can you choose the remaining 2 card from the 39?

OpenStudy (anonymous):

38?

OpenStudy (anonymous):

so 39 * 38 * 37

OpenStudy (zarkon):

no

OpenStudy (zarkon):

this is again a combination

OpenStudy (zarkon):

you are CHOOSING 2 cards from 39

OpenStudy (anonymous):

what where did the 2 come from

OpenStudy (zarkon):

you are picking a 5 card hand. 3 are diamonds...thus the other 2 cards must not be diamonds

OpenStudy (anonymous):

oh psh right :/

OpenStudy (anonymous):

um so after 2, 37's left

OpenStudy (zarkon):

I don't care how many are left after you pick 2. We need to know how many ways we can pick the 2 cards from the 39

OpenStudy (zarkon):

it is the same exact idea we used to choose the 3 diamonds

OpenStudy (anonymous):

39 C 2

OpenStudy (anonymous):

and then multiply it with 13 C 3

OpenStudy (zarkon):

yes

OpenStudy (zarkon):

but that is not the final answer

OpenStudy (zarkon):

because they are asking for AT LEAST 3 diamonds we need to do this tow more times. one with 4 diamonds and one with 5 diamonds

OpenStudy (zarkon):

*two

OpenStudy (zarkon):

N(at least 3 diamonds) =N(get exactly 3 diamonds)+N(get exactly 4 diamonds)+N(get exactly 5 diamonds)

OpenStudy (anonymous):

13 C 4 * 38 C 1; 13 C 5 * 37 C 0

OpenStudy (anonymous):

oh

OpenStudy (zarkon):

almost

OpenStudy (zarkon):

there are always 39 cards that are not diamonds

OpenStudy (anonymous):

LOL shet

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

so it's 13 C 4 * 39 C 1; 13 C 5 * 39 C 0

OpenStudy (zarkon):

yes...so what is the final answer?

OpenStudy (anonymous):

211926?

OpenStudy (zarkon):

no

OpenStudy (anonymous):

oh no

OpenStudy (anonymous):

crap

OpenStudy (anonymous):

one sec

OpenStudy (anonymous):

241098

OpenStudy (zarkon):

yes

OpenStudy (anonymous):

LOL THANK YOU SO MUCH

OpenStudy (zarkon):

no problem

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