Distribution of suits in a 13 card hand Let's say you have 13 cards distributed from a standard deck, find the probability of this distribution of suits: 4, 4, 3, 2, (for instance 4 hearts, 4 clubs, 3 diamonds, 2 spades). My answer was:
$$\frac{ 4*{13\choose 4} * 3*{13\choose 4} * 2*{13\choose 3} * 1 *{13\choose 2} }{52\choose 13}$$
My reasoning being that since there were 4 ways to choose the first suit, 3 ways to to choose the next etc... However the real answer was:
$$\frac{\frac{4!}2 *{13\choose 4} * {13\choose 4} *2 * {13\choose 3} * {13\choose 2}}{52\choose 13}$$
Basically mine divided by 2, what I don't understand why the need to divide by 2. There are 4! ways the suits can be arranged, so why are they arranged in 4!/2 ways in the answer?
 A: You have double-counted. Happens a lot. There is no "first suit" and "second suit." In the $4$ at the beginning of your expression, the choice of $\spadesuit$ was one of the choices counted, and in the subsequent $3$, the choice of $\heartsuit$ is one of the choices counted.
But in the $4$, the choice $\heartsuit$ was counted, and in the subsequent $3$, $\spadesuit$ was one of the choices counted.
But the net result in either case is four each of $\spadesuit$ and $\heartsuit$. 
A way to avoid this is to say that the suits we have four in can be chosen in $\binom{4}{2}$ ways. Once we have done that, the actual cards in the higher ranking suit can be chosen in $\binom{13}{4}$ ways, and then the actual cards in the lower ranking suit can be chosen in $\binom{13}{4}$ ways.
Or else we can deliberately double-count, and then divide by 42$. 
A: 
Find the probability of this distribution of suits: 4, 4, 3, 2.

Order of suit selection doesn't matter. Distinction come from numbers of cards selected from each suit.


*

*Select 2 suits from 4.

*

*Then select 4 of their 13 cards, each.


*Select 1 suit from the 2 remaining. 

*

*Then select 3 of its 13 cards.


*Only 1 suit remains to select.

*

*So select 2 of its 13 cards. 



Multiply the selections together and divide by the total selection space.
$$= \frac{{4\choose 2}{13\choose 4}^2\times{2\choose 1}{13\choose 3}\times{13\choose 2}}{{52\choose 13}}$$

Another approach.  First sort the cards into suits, then arrange the suits into groups, or "boxes", labelled '4','3','2' such that 2 suits can be placed in '4' and 1 suit in each of the others.  The count of ways to so arrange the suits into these groups is given by the multinomial coefficient:
$$\frac{4!}{2!1!1!} = {4\choose 2}{2\choose 1}{1\choose 1}$$
Next, select a number of cards from each suit based on their grouping (the numbers on the boxes' labels) :
$$\frac{13!}{4!9!}\times\frac{13!}{4!9!}\times\frac{13!}{3!10!}\times\frac{13!}{2!11!} = {13\choose 4}^2{13\choose 3}{13\choose 2}$$
Put them together, over the total way to select 13 cards.
$$\frac{\frac{4!}{2!1!1!}{13\choose 4}^2{13\choose 3}{13\choose 2}}{52\choose 13} \\ = \frac{13!^5 39!}{2!^2 3! 4! 9!^2 10! 11! 52!}$$
