A bit of help needed
 

Is this right:

Texas Hold'Em. I'm in the CO with TT. How likely is it that one or more of the 3 people to act after me has JJ, QQ, KK or AA?

I look at it as dealing 6 cards for 50!/44! combinations.

J, J is the first and second card dealt in 4*3*48!/44! of those combinations. Multiply by 4 to get JJ, QQ, KK, and AA. Multiply by 3 for number of players.

So the complete probability is 3*4*4*3*(48!/44!)/(50!/44!) or 5.87755%

I think the result is right. But I was wondering if there's a neater notation for the equation?

I'm also a bit confused about whether the 'standard' C (permutations) function removes order considerations. That is, whether the definition is
C(A, B) = (A! / (A-B)!) or
C(A, B) = (A! / (A-B)!) / B!

Thanks in advance.

Re: A bit of help needed
 

[ QUOTE ]
I'm also a bit confused about whether the 'standard' C (permutations) function removes order considerations.

[/ QUOTE ]
It does. C(N,R) = N!/[(N-R)! * R!]

Working on the first part. Look up the inclusion/exclusion principle to get an idea of how to solve it.