Poker game full house 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. 
 A: 
*

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

A: How many ways could you get a full house of kings over queens? You need 3 of the 4 kings and 2 of the 4 queens. That explains the $4 \choose 3$ and the $4 \choose 2$ factors. But of course you could have a lot of other types of full houses. How many options do you have for the three of a kind? $13 \choose 1$. Then of course you only have 12 other denominations left for the pair - that's the $12 \choose 1$. 
A: There are four suits and 13 cards in each suit. A full house has 3 cards of the same value, and 2 more cards of the same value.
Of the 13 card values choose 1 value for your 3 of a kind: $13 \choose 1$. Now of the 4 suits choose 3 of the cards with this value: $4 \choose 3$.
For the 2 of a kind, there are 12 card values left to choose from (you can't choose the same value from the 3 of a kind, since that would require 5 suits): $12 \choose 1$. Now of the 4 suits choose 2 of the cards with this value: $4 \choose 2$.
Multiply them all together to get the number of combinations.
