What is the probability of getting a 5 card hand where the 5 cards are from exactly 2 suits
nCr = n!/(r!(n-r)!)
The Attempt at a Solution
There are two possible cases where the 5 cards are from 2 suits, case one is 4 are from one suite and 1 is from another and case two is that 3 are from one sweet and 2 are from another.
there are 4 ways to choose a suit and then 13C4 ways to choose 4 cards from that suit. Then there are 39 ways to choose the last card because we must pick from the remaining 3 suites
There are 4 ways to choose a suit and then 13C3 ways to pick 2 cards from it. Then there are 3 ways to pick another suit and 13C2 ways to pick 2 cards from it
since the total number of 5 card hands is 52C5 the probability where the 5 are from 2 suits is
[4(13C4)39 + 4(13C3)3(13C2)]/52C5
is this correct?