Mathematics
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OpenStudy (anonymous):
From a deck of 52 cards, how many 5-card hands can be formed with at least 3 diamonds?
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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?
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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
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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
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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
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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
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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
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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
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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
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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?
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OpenStudy (zarkon):
no
OpenStudy (anonymous):
oh no
OpenStudy (anonymous):
crap
OpenStudy (anonymous):
one sec
OpenStudy (anonymous):
241098
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OpenStudy (zarkon):
yes
OpenStudy (anonymous):
LOL THANK YOU SO MUCH
OpenStudy (zarkon):
no problem