Question

How many poker hands (of 5 total cards) are dealt such that the first three cards have

the same rank, but the total hand does not contain another pair and is not a four of

a kind? When counting for this problem, the order of the dealt cards matters.

plz explain your solution

Answer 1

Well we jave total 13 different rank in our pack of 52 cards so for selecting 3 cards of same rank, first we have to select a rank and then 3 card from this selected rank.In this way we always select 3 cards from 3 different kind ( we have total 4 kind in our whole pack known as spade,club,heart and daimond)

So we have no issue with 4 of same kind bacause rest 2 cards can be of any kind but not from the same rank then we have cards from atleast 3 ranks hence cannot be 4 of same kind.(If we are choosing 5 card and we need 4 of same kind then we can choose at most two kind )

Hence number of ways to selact a rank is C(13,1) =13

And out of this selected rank 3 card can be choosed in C(4,3)=4 ways now rest 2 card can be any out of rest 48 cards by excluding the fourth card of same rank hence that can be done in C(48,2)=1128 ways

Hence total number of ways of selecting such 5 card is

=13×4×1128=58656