Quoting the book I'm reading:
A poker hand consists of 5 cards. If the cards have distinct consecutive values and are not all of the same suits, we say that the hand is a straight. For instance, a hand consisting of the five of spades, six of spades, seven of spades, eight of spades, and nine of hearts is a straight. What is the probability that one is dealt a straight?
The book gives a solution to this problem, but I don't quite agree with the combinatorics of it. It starts calculating how many poker hands there are, it argues that there are $52\choose 5$, which I totally agree since a hand is to choose 5 cards in 52 where the order doesn't matter, so it's not $52\times51\times50\times49\times48$. Now for the part I don't agree, after it the book reads:
To determine the number of outcomes that are straights, let us first determine the number of possible outcomes for which the poker hand consists of an ace, two, three, four, and five (the suits being irrelevant). Since the ace can be any 1 of the 4 possible aces, and similarly for the two, three, four, and five, it follows that there are $4^5$ outcomes leading to exactly one ace, two, three, four, and five.
My question is since the order doesn't matter a hand like A-2-3-4-5 and A-4-3-5-2 are both straights and are the same hand, actually, all the $5!$ permutations of it are the same hand, so there should be (ignoring, for now, the suit) $(4^5)/5!$ unique straight hands that the lower card is an ace! Now, considering the suit, we have to subtract 4 hands, which the book does. Where is my reasoning wrong?