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$$4 \choose 2 Can someone explain me in details why we multiply those combinations? I mean, explain why we consider these numbers. 3 Answers • First we select which type's 3 cards we want: There are 13 types(or ranks) of cards: A,K,Q,J,10,9,8,7,6,5,4,3,2,1.We select one of them.$$\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$$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$12 \choose 1$. 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.