[Math] Poker game full house

Question

I'm dealing with an exercise which deals with the poker game. I need to calculate the probability of getting a full house.

Full house is getting 3 cards of the same type and 2 cards of the same type.

I've made a research, but I cannot understand why the combination for getting a full house is

$13 \choose 1$$4 \choose 3$$12 \choose 1$$4 \choose 2$
Can someone explain me in details why we multiply those combinations? I mean, explain why we consider these numbers.

Answer 1

• First we select which type's 3 cards we want:

  There are 13 types(or ranks) of cards: $A,K,Q,J,10,9,8,7,6,5,4,3,2,1$.We select one of them.

$$\binom{13}1$$

• Select three cards from it:

  There are only four cards of different suites of same type: $K\heartsuit,K\diamondsuit,K\spadesuit,K\clubsuit$, we select any three of them

$$\binom{4}3$$

• Select second type who's 2 cards you want:

  From the remaining $13-1=12$ types we select one type.

$$\binom{12}1$$

• Select two cards out of it:

  Similarly we have $4$ suits, we select two of them of a single type.

$$\binom{4}2$$

• So total ways:

  Product of all ways, multiplication theorem.

$$\binom{13}1*\binom{4}3*\binom{12}1*\binom{4}2$$