How many combinations of five card hands are there in a 52 card deck
where three will be of one suit and two of one other (ie harts and
clubs) , as well how can one ascertain the probability of this
outcome?

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Clarification of Question by hamstersproblem-ga on 24 Oct 2005 07:23 PDT

is this the same if we are specifying the two suits "up front"?

We approach this problem by breaking it down into steps.
I will use notation like 5P2 and 5C2 to indicate permutations and
combinations respectively.

a). How many ways can we pick the two suits?
 - We can pick any suit for the three cards, and any different suit
for the two cards. Order is important - 3H and 2C is not the same as
3C and 2H. So the answer is 4 * 3 (or 4P3) = 12.
b). How many ways can we pick three cards in the specified suit?
 - We want to select three cards from the 13 in the given suit. Order
is not important. There are 13C3 = 13*12*11/(3*2*1 ) = 286 ways.
c). How many ways can we pick the two cards in the other suit?
 - Similar to b), but we only want two cards now. There are 13C2 =
13*12/2 = 78 ways.

Putting these together, there are 12 * 286 * 78 = 267,696 ways.

The probability of this or any other outcome is defined as the number
of ways we can get it divided by the total number of possible hands,
which is 52C5 = 2,598,960. In this case the probability is 267,696 /
2,598,960 = 0.103 or 10.3%.

sorry, in (a) should be 4P2, not 4P3.

is this the same if we are specifying the two suits "up front"?