Answer this problem

[Stl Lunatic Profile WebsiteYIM]

How many 10-card hands exist, where 6 are one suit and 4 are another suit?

Use probability equations if you know how...N= # of posibilities in first example
R=number of examples

Ex. _ _ _ _ _ mean r=5, first example is the first _

Exponential=N(to the power of r)
Repitition
Order matters

Permutation=N!/(N-R)!
no repitition
order matters

Combonation=N!/(N-R)!/R!
no repitition
order doesnt matter


If anyone knows how to do this it would be great...I think it might be a trick problem so I'm not sure...Good Luck

Thanks

[gowhitesox99 Profile]

awww damms.... 4 years ago i could have hooked you up Luna! I have forgotten all that crap....

[Stl Lunatic Profile WebsiteYIM]

Yeah, this problem has been irking me for a while. It's different than the normal problems I'v been doing. I think you are soposed to use the combonation method to figure it out. I don't think there can be very many posibilities...i think you have to use half of 52 somehow which is 26 because you are using only 2 of the four suits...6/13 and 4/13 of each suit...then maybe this is a 10-13 problem...

like the N=13 and R=10

That gives me 286 possibilities...but then I really don't know if that works out...

Does this ring any bells for anyone lol?

Its confusing because you can only have 26 or half of the cards so you cant use the number 52 because there is no chance of using anything above 50% of 52 cards...so you get 26 which then can be split into two scenerios of 13 cards each...for some reason I think I figured it out and that is the answer (286)

[TChicken Profile]

286 is way to low, think of it more as several smaller comb/perm problems combined.