First off, this is a combinations question - we don't care about the order in which the cards are dealt. The general formula for combinations is:
#C_(n,k)=((n),(k))=(n!)/((k!)(n-k)!)# with #n="population", k="picks"#
Before moving on, let's see how many 5 card hands are possible:
#C_(52,5)=((52),(5))=(52!)/((5)!(52-5)!)=(52!)/((5!)(47!))#
Let's evaluate it!
#(52xx51xxcancelcolor(orange)(50)^10xx49xxcancelcolor(red)48^2xxcancelcolor(brown)(47!))/(cancelcolor(orange)5xxcancelcolor(red)(4xx3xx2)xxcancelcolor(brown)(47!))=52xx51xx10xx49xx2=2,598,960#
So of those nearly 2.6 million hands, how many are 2 pair hands?
To achieve 2 pair, we first need to select, from the 13 ordinals (Ace through 10, Jack, Queen, King) 2 of them:
#((13),(2))#
From each of those ordinals, we want 2 suits from the four available (spades, hearts, diamonds, clubs) each:
#((13),(2))((4),(2))^2#
And now we need a last card. It needs to be one of the remaining 11 ordinals and we'll be choosing one card:
#((13),(2))((4),(2))^2((11),(1))((4),(1))=78xx6^2xx11xx4=123,552#
A helpful site for doing poker hand probability and combinametrics is //en.wikipedia.org/wiki/Poker_probability
The question is:
In a game of poker, five players are each dealt $5$ cards from a $52$-card deck. How many ways are there to deal the cards?
The answer is $\frac{52P5}{(5!)^5}$ or $\frac{52!}{47!5!}$ I understand the reason after seeing the answer: $25$ ordered cards are picked from the deck of $52$, then divided by the number of ways each hand can be ordered, because the order of a hand does not matter.
However, before seeing the answer I came up with this attempt:
You choose $5$ cards from $52$, the order doesn't matter so we have $52\choose{5}$. Then you're left with $47$ cards and for the next hand we have $47\choose{5}$. The final sum is ${{52}\choose{5}} + {{47}\choose{5}} + {{42}\choose{5}} + {{37}\choose{5}} + {{32}\choose{5}}$.
This sum is considerably larger, but while I understand the reason $\frac{52!}{47!5!}$ is the correct, I can't find the mistake in my first attempt.