combinatorics
Can someone explain intuitively why this is 52 C 5? What does it mean that the "order doesn't matter?" If I wanted to pick 5 cards from a 52 card deck, why wouldn't I just do 52!/47! since there are 52 spots for the first card, 51 for the second, 50 for the third, etc.?
Any help would be greatly appreciated!
Your reasoning was no so evident for me at first, so I decided to dig a little and what I first found was a table with an entry that agrees with your interpretation, but then I wanted to know where that multichoose formula comes from, and after a while I found useful this explanation using the bars and stars approach.
In your case, for example, we would represent the $10$ cards of the hand as "stars" and then we would put $52-1$ "bars" between them to represent the idea of $52$ different types$^1$ of cards. The stars that are to the left of a bar are of the same type. We have then to be careful with the extreme case where all the $10$ stars are to the left of a bar, meaning that the all the $10$ cards are all of the same type. As you and @JMoravitz correctly pointed out, this is possible because the size of the hand is not greater than the multiplicity of each type in the superdeck.
Under this approach we could imagine that we have a total of $52+10-1$ bins to place the stars and the bars, and then ask in how many ways we could place the $10$ stars out of a total $52+10-1$ available bins. Now the answer is more evident to me, in
$$\binom{52+10-1}{10}$$
ways. Once you place the stars there is only one way to put the bars, since the order does not matter.
EDIT: I was wondering why Bose-Einstein would be a hint in the problem. Following the perspective proposed in this video, it seems that this problem is analogous to the Bose-Einstein statistics in that we could thought of the $10$ cards (stars) in the hand as indistinguishable particles and the $52^*$ types of cards in a deck as energy states where those particles could condensate (don't get me wrong, I am not a quantum physicist). In any case, it's not clear how that would be a hint to solve the problem...any hint?
$^*$ Thanks to Michael's answer I understood that the number of energy states is $52$ and not $52+10-1$, as I originally wrote in my edit...although it still seems a rather sophisticated hint for me.
$^1$ Here the type of a card is its rank and suit.
Break into two cases:
For ease of calculation, we treat the cards in our hand as order not mattering, allowing us to use binomial coefficients and not have to break this case down further into aaabc, aaacb, aabac, aabca, etc...
In trying to calculate the number of outcomes in the first case, we continue:
Multiplying these out give the number of hands where order doesn't matter which have one suit that has three cards and two suits which each have one card as being $$4\times \binom{3}{2}\times \binom{13}{3}\times 13\times 13$$
Now, if you want order of cards to matter, just multiply the result by $5!$.
The second case is calculated similarly, and I leave it to you to complete yourself.
Edit: More details
When applying multiplication principle, outcomes selected in one step are treated differently than outcomes selected in a later step. As such, in order to avoid overcounting, you need to ask yourself the question of if you were to answer the steps in one order you could have given answers in a different order and still arrived at the same outcome. Similarly, to avoid undercounting you need to ask yourself if every outcome is actually counted.
To illustrate your mistake here, let us consider a smaller problem. We have four people, two men and two women. Let the women be $A,B$ and the men be $X,Y$. We ask the question of "how many ways are there to select four people such that both genders are represented?"
Here... it should be painfully obvious that every way of selecting the people will have both genders represented. If we were to treat the selections as order matters, then this is very simply $4!$ number of arrangements, and if we treat selections as order doesn't matter there is very simply only $1$ outcome.
According to your logic, however... "Pick a representative for a first gender, pick a representative for the remaining gender, then pick the remaining" when counting the number of arrangements where order matters you would have had an answer along the lines of $4\times 2\times 2\times 1$ which is not the full $4!$. The reason why is that you very specifically counted only the cases where the first two people in the line are of a different gender and neglected to consider outcomes like $XYAB$ and $ABXY$, etc...
On the other hand, if we treated order as not being important, according to your logic, "Pick the used genders, pick a guaranteed representative for the males, pick a guaranteed representative for the females, pick the remaining people" you would have arrived at an answer similar to $\binom{2}{2}\times 2\times 2\times \binom{2}{2}=4$ which is certainly not $1$ which we knew ahead of time to be the correct answer. The error here is that by selecting a person to be the guaranteed representative for a gender, we have placed unnecessary importance on them. This would have been the correct answer to the question of picking four people from our four people, such that there are two genders present, where each gender has a designated "leader." Compare the proposed steps with "Pick the used genders, pick a leader for the males, pick a leader for the females, pick the remaining people."
To correctly count... anything which can be confused for one another if they were picked in sequence, instead pick simultaneously. In the first example, looking at a final outcome there is no confusion as to which suit was the suit that has three cards in it, it is obvious which that is. The two suits which each have one card however, we wouldn't have been able to tell which suit was chosen first and which was chosen second had the cards been mixed up. In the same way, the three cards chosen for the suit with three cards in it, if we were to pick them in sequence and then mixed the cards, we won't have the ability to know in what order they were picked any more after the mixing.
The steps I would use for correctly counting the second case, which again is the number of ways in which we can have two suits with two cards each and one suit with one card such that order doesn't matter:
This gives $\binom{4}{2}\binom{13}{2}\binom{13}{2}\times 26$ outcomes for the second case.
The final count where order doesn't matter is then:
$$4\times \binom{3}{2}\times\binom{13}{3}\times 13\times 13 + \binom{4}{2}\binom{13}{2}\binom{13}{2}\times 26$$
Where order does matter, multiply the above result by $5!$
Best Answer
Your calculation treats $\heartsuit 3,\spadesuit 4,\clubsuit 5,\diamondsuit 6,\heartsuit 7$ and $\diamondsuit 6,\spadesuit 4,\heartsuit 7,\heartsuit 3\clubsuit 5$, for instance, as different hands, when in fact they’re the same set of five cards: the only difference between them is the order in which they were dealt. Since a given set of $5$ cards can be dealt in $5!$ different orders, you’re counting each hand of $5$ cards $5!$ times. To get rid of this overcounting, you must divide by $5!$. When you do, you get
$$\frac{52!}{47!}\cdot\frac1{5!}=\frac{52!}{5!47!}=\binom{52}5\;,$$
the number of ways of choosing an unordered set of $5$ cards from a $52$-card deck. Your $\dfrac{52!}{47!}$ is the number of ways to deal $5$ cards: it counts each of the $5!=120$ possible dealing orders of a given hand separately.