- anonymous

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

- schrodinger

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- anonymous

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

- anonymous

wut.

- Zarkon

A lot.
What have you tried?

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## More answers

- anonymous

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

- Zarkon

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

- anonymous

:/ no...

- anonymous

5 hand?

- Zarkon

how many ways can you choose 3 cards from 13?

- anonymous

so 13 Cr 3?

- anonymous

o wait

- anonymous

i meant 13 nCr 3

- Zarkon

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

- Zarkon

ok
how many cards are not diamonds?

- anonymous

i don't know

- Zarkon

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

- anonymous

39

- Zarkon

ok

- Zarkon

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

- anonymous

38?

- anonymous

so 39 * 38 * 37

- Zarkon

no

- Zarkon

this is again a combination

- Zarkon

you are CHOOSING 2 cards from 39

- anonymous

what where did the 2 come from

- Zarkon

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

- anonymous

oh psh right :/

- anonymous

um so after 2, 37's left

- 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

- Zarkon

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

- anonymous

39 C 2

- anonymous

and then multiply it with 13 C 3

- Zarkon

yes

- Zarkon

but that is not the final answer

- 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

- Zarkon

*two

- Zarkon

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

- anonymous

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

- anonymous

oh

- Zarkon

almost

- Zarkon

there are always 39 cards that are not diamonds

- anonymous

LOL shet

- anonymous

ok

- anonymous

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

- Zarkon

yes...so what is the final answer?

- anonymous

211926?

- Zarkon

no

- anonymous

oh no

- anonymous

crap

- anonymous

one sec

- anonymous

241098

- Zarkon

yes

- anonymous

LOL THANK YOU SO MUCH

- Zarkon

no problem

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