Question

In 5-card hand poker, how many full houses exists where the two-pair are Jacks? A full house is three-of-a-kind and two-of-a-kind, so in this question, the two-of-a-kind must be two Jacks.

Answer 1

In this case since 2 jacks are already present , you just need to find the remaining 3 cards of the same kind

Now since jack is already taken , there are 12 cards each of 4 different suits , now for similar kind you have to select three out of those 4 different suit of a single Card i.e in 4C3 = 4 ways you can select a single Card of different suit.

Take for example 1

You can select

1Heart 1spade 1diamond

1Heart 1Spade 1Club

1Heart 1Club 1Diamond

1Spade 1Club 1Diamond , hence 4 ways

Now there are 12 different cards ( excluding jack)

So therefore total number of ways = 12*4 = 48

Which is the required answer