# About choosing suits and ranks in five hand poker

Full House: $\binom{13}1 \binom 43 \binom{12}1 \binom 42$

One Pair: $\binom{13}1 \binom 42 \binom{12}3 \binom 41^3$

Two Pair: $\binom{13}2 \binom 42^2 \binom {44}1$

How do they know how many subsets to choose?

In Full House they choose two ranks as $\binom{13}1 \binom{12}1$. How do they know it's not $\binom{13}2?$

In Two Pair they choose ranks as $\binom{13}2.$ Why not $\binom{13}1 \binom {12}1$ or $\binom{13}1 \binom {13}1$?

Also, why is $\binom43$ in Full House not $\binom41^3$ and conversely why is $\binom41^3$ in One Pair not $\binom43$?

I am having hard time seeing how they decide on these things. Is there a general rule that I could use?

In Full House they choose two ranks as ${13\choose 1}{12\choose 1}$ vs. ${13\choose 2}.$

Because Aces over 8's is fundamentally different from 8's over A's

"In Two-Pair they choose two ranks as ${13\choose 2}$ v. ${13\choose 1}{12\choose 1}.$"

This doesn't have that problem. Pairs of Aces and 8's are the same as pairs of 8's and A's.

Why is ${4\choose 3}$ in Full House not ${4\choose 1}^3$

${4\choose 3}$ says: "There are 4 objects in a set. Choose 3 of them." Or, 3 cards of the same rank.

${4\choose 1}^3$ says: "there are 3 sets each with 4 objects. Choose one object from each set."

In Full House they choose two ranks as $\binom{13}{1}\binom{12}{1}$. How do they know it's not $\binom{13}{2}$?

Unlike Two Pair values choosen are different in sense that the first one gives three cards and the second one gives only two cards.

In Two Pair they choose ranks as $\binom{13}2.$ Why not $\binom{13}1 \binom {12}1$ or $\binom{13}1 \binom {13}1$?

Unlike Full House values choosen are similar, because they both give two cards.

Also, why is $\binom43$ in Full House not $\binom41^3$ and conversely why is $\binom41^3$ in One Pair not $\binom43$?

Here $\binom43$ for Full House gives number of ways to choose 3 cards of 4 with the same value. One Pair doesn't give any restriction on choosing suits for single cards of different values, because we alreade defined values to make these cards different, so there are 3 independent choices.

So general rule is the following. Think how many objects (suits, values, cards, anything else) you have and how many similar of them you would like to choose. Use the nunber of chooses here. Multiply independent choices. Think more for dependent case.

Full House: $\binom{13}1 \binom 43 \binom{12}1 \binom 42$

In Full House they choose two ranks as $\binom{13}1\binom{12}1$. How do they know it's not $\binom{13}2$?

Because we are selecting cards for distinct things: a pair and a triple. eg: "Two kings and three queens" is distinct from "Two queens and three kings".

We wish to obtain three from four suits for one from thirteen kinds, and two from four suits for one from twelve other kinds.

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

Two Pair: $\binom{13}2 \binom 42^2 \binom {44}1$

In Two Pair they choose ranks as $\binom{13}2.$ Why not $\binom{13}1 \binom {12}1$ or $\binom{13}1 \binom {13}1$?

Because we a selecting cards for indistinct things; eg: "Two kings and two queens" is indistinguishable from "Two queens and two kings"

So we don't worry about order of selection to the pairs.

We wish to obtain two suits for each of two kinds, and one card from the remaining deck (44 cards, that is: one suit for one from the 11 other kinds).

$$\binom 42^2\binom {13}2\cdot \binom{44}{1} ~=~ \binom 42^2\binom {13}2\cdot \binom{4}{1}\binom{11}1$$

One Pair: $\binom{13}1 \binom 42 \binom{12}3 \binom 41^3$

Also, why is $\binom43$ in Full House not $\binom41^3$ and conversely why is $\binom41^3$ in One Pair not $\binom43$?

Because we are selecting three suits for one kind in the full house, and one suit for each of three kinds in the one pair senario.

We wish to obtain two from four suits for one from 13 kinds, and one from four suit for each of three from 12 other kinds.

$$\binom{4}2\binom{13}1\cdot \binom{4}1^3\binom{12}3$$