What is the probability that a ﬁve-card poker hand contains cards of ﬁve different kinds? The textbook also states the following: 

There are $13$ different kinds of cards, with four cards of each kind.
  (Among the terms commonly used instead of “kind” are “rank,” “face
  value,” “denomination,” and “value.”) These kinds are twos, threes,
  fours, fives, sixes, sevens, eights, nines, tens, jacks, queens,
  kings, and aces. There are also four suits: spades, clubs, hearts, and
  diamonds, each containing 13 cards, with one card of each kind in a
  suit.

which describes it what it means by a 'kind'.
So far I did the following:
$$\Large\frac{\binom{13}{5}}{\binom{52}{5}}$$
I think I am missing something in the numerator however. Maybe $\binom{47}{5}$? But I am not sure how to justify it.
Edit: Had to edit the question, so if somebody already started answering please check it.
 A: We must select five of the $13$ kinds, which can be done in $\binom{13}{5}$ ways.  For each of the five kinds we can select, there are four suits from which we can select a card of that kind.  Hence, the number of ways we can select a hand in which each card is of a different kind is 
$$\binom{13}{5} \cdot 4^5$$
Hence, the probability that a five card poker hand contains cards of five different kinds is 
$$\frac{\binom{13}{5} \cdot 4^5}{\binom{52}{5}}$$
A: You have $52$ choices for the first card. It remains $52-4$ cards of a different kind for the second one, $52-4\cdot2$ for the third, $52-4\cdot 3$ for the fourth and $52-4\cdot 4$ for the fifth. It gives
$$(52-4\cdot 0)(52-4\cdot 1)(52-4\cdot 2)(52-4\cdot 3)(52-4\cdot 4)$$
But you have to discount the permutations of $5$ cards, that is $5!$. It gives
$$\frac{(52-4\cdot 0)(52-4\cdot 1)(52-4\cdot 2)(52-4\cdot 3)(52-4\cdot 4)}{5!}$$
possibilities. The probability is then
$$\frac{\frac{(52-4\cdot 0)(52-4\cdot 1)(52-4\cdot 2)(52-4\cdot 3)(52-4\cdot 4)}{5!}}{52\choose 5}\approx 0.507$$
A: Pr = [Choose ranks * Choose suit for each chosen rank ] / [Choose any 5 cards]
= $\dfrac{\dbinom{13}5{\dbinom41}^5}{\dbinom{52}5}$
A: Suppose the first card dealt is an Ace. Then of the 51 cards remaining in the deck, 3 are Aces, 48 are not Aces. From these 51 cards, and for our second deal, we require one card from out of the 48 cards that are not Aces. The chance for this event to occur is 48/51.
Assuming the second card dealt is not an Ace, we now have 50 cards in the deck of which 3 are Aces and 3 have the same face value as the second card just dealt. There remain 50-(3+3)=50-6=44 cards which have different face values from the first two cards dealt, and we require one card from these 44 for our next deal. The chance for this event to occur is 44/50.
Proceeding in this way, the probability we are seeking is given by
(48/51)(44/50)(40/49)*(36/48) which is approximately .5071
This is the correct answer, and appears in the book "An Introduction To Probability" by William Feller, in the section titled 'Subpopulations And Partitions'.
