# How many different orderings of the given 8 cards do we obtain by the specified shuffling?

Suppose we have deck of 8 different playing cards, which we shuffle in the following manner :

1. First, we split the deck to halves.
2. Then, we place the cards of the second half between cards of the first half.

So that: the first card of the second half follows the first card of the first half, the second card of the second half follows the second card of the first half, etc.

We repeat this shuffling over and over again.

Question:

1. How many different orderings of the given 8 cards do we obtain by the specified shuffling?
2. Compare the result to the number of all possible orderings of the given 8 cards.

What i have done so far :

So if we split 8 cards into 2 decks with 4 cards each deck we will have 4! to ordering the first decks and 4! to ordering the second decks. So the answer is 4! * 4!.

I so confuse with this problem and above is what I can do. I fell it have something wrong but I do not know where. Please show step if anyone know how to answer this problem.

We start with $$H = \{1,2,3,4,5,6,7,8\}$$

Now we split them in half:

$$H_1 = \{1,2,3,4\}, \space H_2 = \{5,6,7,8\}$$

After placing cards of the second half between cards of the first half:

$$H = \{1,5,2,6,3,7,4,8\}$$

Now we split them in half again:

$$H_1 = \{1,5,2,6\},\space H_2=\{3,7,4,8\}$$

After placing cards of the second half between cards of the first half:

$$H = \{1,3,5,7,2,4,6,8\}$$

Now we split them in half again:

$$H_1 = \{1,3,5,7\},\space H_2 = \{2,4,6,8\}$$

After placing cards of the second half between cards of the first half:

$$H = \{1,2,3,4,5,6,7,8\}$$

We're back to where we started, so we count all the $$H$$ sets prior to this point (there are three of them).

There are $$8! = 40320$$ possible orderings of the $$8$$ cards, so we can conclude that cycling through $$3$$ distinct orderings is a very bad way to shuffle cards. That's probably good to know if you want to perform magic tricks.