What is the number of five-card hands possible from a standard fifty-two card deck? 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!
 A: 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.
A: If a set has $n$ elements, the number of $k$ element subsets is ${n\choose k}$.  You can prove that 
$${n\choose k} = {n!\over k!(n-k)!}\qquad 0\le k \le n.$$
A: The reason you wouldn't do $52!/47!$ is that getting the five of hearts first and the three of clubs second is the same as getting the three of clubs first and the five of hearts second --- either way, you get the same two cards, so you shouldn't count it as two different hands. Instead of $52\times51$, you have to divide that by 2. Same thing, only more so, when dealing a 5-card hand. 
